In Cartesian geometry in two dimensions the
p
a
r
a
b
o
l
a
{\displaystyle parabola}
is the locus of a point that moves so that it is always equidistant from a fixed point and a fixed line.
The fixed point is called the
f
o
c
u
s
{\displaystyle focus}
and the fixed line is called the
d
i
r
e
c
t
r
i
x
{\displaystyle directrix}
. Distance from
f
o
c
u
s
{\displaystyle focus}
to
d
i
r
e
c
t
r
i
x
{\displaystyle directrix}
is non-zero.
Figure 1: The Parabola
y
=
x
2
4
q
{\displaystyle y={\frac {x^{2}}{4q}}}
Focus at point
F
(
0
,
q
)
{\displaystyle F\ (0,q)}
Vertex at origin
(
0
,
0
)
{\displaystyle (0,0)}
Directrix is line
y
=
−
q
{\displaystyle y=-q}
By definition
P
1
F
=
P
1
D
1
,
P
2
F
=
P
2
D
2
,
P
4
F
=
P
4
D
4
.
{\displaystyle P_{1}F=P_{1}D_{1},\ P_{2}F=P_{2}D_{2},\ P_{4}F=P_{4}D_{4}.}
V
F
=
V
D
3
=
q
.
{\displaystyle VF=VD_{3}=q.}
In Figure 1
q
=
1
{\displaystyle q=1}
See Figure 1.
The focus is point
F
(
0
,
q
)
{\displaystyle F\ (0,q)}
and the directrix is line
D
1
D
4
{\displaystyle D_{1}D_{4}}
:
y
=
−
q
.
{\displaystyle :\ y=-q.}
The
v
e
r
t
e
x
{\displaystyle vertex}
, point
V
{\displaystyle V}
:
(
0
,
0
)
{\displaystyle :\ (0,0)}
, is half-way between
focus and directrix. A
c
h
o
r
d
{\displaystyle chord}
is the segment of a line joining any two distinct points of the parabola. The line segment
P
2
F
P
4
{\displaystyle P_{2}FP_{4}}
is a chord. Because chord
P
2
F
P
4
{\displaystyle P_{2}FP_{4}}
passes through the focus
F
{\displaystyle F}
, it is called a
f
o
c
a
l
c
h
o
r
d
.
{\displaystyle focal\ chord.}
The focal chord parallel to the directrix is called the
l
a
t
u
s
r
e
c
t
u
m
.
{\displaystyle latus\ rectum.}
The line through the focus and perpendicular to the directrix is the
a
x
i
s
{\displaystyle axis}
, sometimes called
a
x
i
s
o
f
s
y
m
m
e
t
r
y
{\displaystyle axis\ of\ symmetry}
.
Let an arbitrary point on the curve be
(
x
,
y
)
{\displaystyle (x,y)}
.
By definition,
(
x
−
0
)
2
+
(
y
−
q
)
2
=
y
+
q
{\displaystyle {\sqrt {(x-0)^{2}+(y-q)^{2}}}=y+q}
. This expression expanded gives:
x
2
−
4
q
y
=
0
;
y
=
x
2
4
q
{\displaystyle x^{2}-4qy=0;\ y={\frac {x^{2}}{4q}}}
.
If the equation of the curve is expressed as:
y
=
K
x
2
{\displaystyle y=Kx^{2}}
, then
K
=
1
4
q
;
4
K
q
=
1
;
q
=
1
4
K
{\displaystyle K={\frac {1}{4q}};\ 4Kq=1;\ q={\frac {1}{4K}}}
where the
f
o
c
u
s
{\displaystyle focus}
has coordinates
(
0
,
q
)
,
{\displaystyle (0,q),}
and
q
{\displaystyle q}
is the distance form vertex to focus.
If the directrix is parallel to the
X
{\displaystyle X}
axis, then the parabola is the same as the familiar quadratic function.
The general parabola allows for a directrix anywhere with any orientation.
The General Parabola
edit
Let the directrix be
a
x
+
b
y
+
c
=
0
{\displaystyle ax+by+c=0}
where at least one of
a
,
b
{\displaystyle a,b}
is non-zero.
Let the focus be
(
p
,
q
)
{\displaystyle (p,q)}
.
Let
(
x
,
y
)
{\displaystyle (x,y)}
be any point on the curve.
Distance from point
(
x
,
y
)
{\displaystyle (x,y)}
to focus
(
p
,
q
)
{\displaystyle (p,q)}
=
(
x
−
p
)
2
+
(
y
−
q
)
2
{\displaystyle {\sqrt {(x-p)^{2}+(y-q)^{2}}}}
.
Distance from point
(
x
,
y
)
{\displaystyle (x,y)}
to directrix (
a
x
+
b
y
+
c
=
0
{\displaystyle ax+by+c=0}
)
=
a
x
+
b
y
+
c
R
{\displaystyle {\frac {ax+by+c}{\sqrt {R}}}}
where
R
=
a
2
+
b
2
{\displaystyle R=a^{2}+b^{2}}
.
By definition these two lengths are equal:
(
x
−
p
)
2
+
(
y
−
q
)
2
=
a
x
+
b
y
+
c
R
{\displaystyle {\sqrt {(x-p)^{2}+(y-q)^{2}}}={\frac {ax+by+c}{\sqrt {R}}}}
.
∴
R
(
x
−
p
)
2
+
(
y
−
q
)
2
=
a
x
+
b
y
+
c
{\displaystyle \therefore \ {\sqrt {R}}{\sqrt {(x-p)^{2}+(y-q)^{2}}}=ax+by+c}
.
Square both sides:
R
(
(
x
−
p
)
2
+
(
y
−
q
)
2
)
=
(
a
x
+
b
y
+
c
)
2
{\displaystyle R((x-p)^{2}+(y-q)^{2})=(ax+by+c)^{2}}
.
R
(
(
x
−
p
)
2
+
(
y
−
q
)
2
)
−
(
a
x
+
b
y
+
c
)
2
=
0
{\displaystyle R((x-p)^{2}+(y-q)^{2})-(ax+by+c)^{2}=0}
.
Expand and the result is:
b
2
x
2
−
2
a
b
x
y
+
a
2
y
2
−
2
(
a
c
+
p
R
)
x
−
2
(
b
c
+
q
R
)
y
+
R
(
p
2
+
q
2
)
−
c
2
=
0
…
(
1
)
{\displaystyle b^{2}x^{2}-2abxy+a^{2}y^{2}-2(ac+pR)x-2(bc+qR)y+R(p^{2}+q^{2})-c^{2}=0\ \dots \ (1)}
.
(
1
)
{\displaystyle (1)}
has the form of the equation of the conic section
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}
where
A
=
b
2
B
=
−
2
a
b
C
=
a
2
D
=
−
2
(
a
c
+
p
R
)
E
=
−
2
(
b
c
+
q
R
)
F
=
R
(
p
2
+
q
2
)
−
c
2
B
2
−
4
A
C
=
0
R
=
a
2
+
b
2
{\displaystyle {\begin{aligned}A&=b^{2}\\B&=-2ab\\C&=a^{2}\\D&=-2(ac+pR)\\E&=-2(bc+qR)\\F&=R(p^{2}+q^{2})-c^{2}\\\\B&^{2}-4AC=0\\R&=a^{2}+b^{2}\end{aligned}}}
B
2
=
4
A
C
{\displaystyle B^{2}=4AC}
because this curve is a parabola.
Figure 2: The Parabola
16
x
2
+
24
x
y
+
9
y
2
−
20
x
−
140
y
+
100
=
0
{\displaystyle 16x^{2}+24xy+9y^{2}-20x-140y+100=0}
Green line is
d
i
r
e
c
t
r
i
x
:
3
x
−
4
y
−
5
=
0
{\displaystyle directrix:\ 3x-4y-5=0}
Blue line is
a
x
i
s
:
4
x
+
3
y
−
10
=
0
{\displaystyle axis:\ 4x+3y-10=0}
Focus at point
F
:
(
1
,
2
)
{\displaystyle F:\ (1,2)}
Vertex at point
V
:
(
1.6
,
1.2
)
{\displaystyle V:\ (1.6,1.2)}
V
F
=
1.
{\displaystyle VF=1.}
Shape of curve is:
y
=
x
2
4
.
{\displaystyle y={\frac {x^{2}}{4}}.}
See Figure 2.
(
a
,
b
,
c
)
=
(
3
,
−
4
,
−
5
)
{\displaystyle (a,b,c)=(3,-4,-5)}
(
p
,
q
)
=
(
1
,
2
)
{\displaystyle (p,q)=(1,2)}
The equation of the parabola is derived as follows:
A
=
b
2
=
(
−
4
)
2
=
16
B
=
−
2
a
b
=
−
2
(
3
)
(
−
4
)
=
24
C
=
a
2
=
(
3
)
2
=
9
R
=
a
2
+
b
2
=
25
D
=
−
2
(
a
c
+
p
R
)
=
−
2
(
(
3
)
(
−
5
)
+
(
1
)
(
25
)
)
=
−
2
(
−
15
+
25
)
=
−
2
(
10
)
=
−
20
E
=
−
2
(
b
c
+
q
R
)
=
−
2
(
(
−
4
)
(
−
5
)
+
(
2
)
(
25
)
)
=
−
2
(
20
+
50
)
=
−
2
(
70
)
=
−
140
F
=
R
(
p
2
+
q
2
)
−
c
2
=
25
(
1
+
4
)
−
25
=
100
{\displaystyle {\begin{aligned}&A=b^{2}=(-4)^{2}=16\\&B=-2ab=-2(3)(-4)=24\\&C=a^{2}=(3)^{2}=9\\&R=a^{2}+b^{2}=25\\&D=-2(ac+pR)=-2((3)(-5)+(1)(25))=-2(-15+25)=-2(10)=-20\\&E=-2(bc+qR)=-2((-4)(-5)+(2)(25))=-2(20+50)=-2(70)=-140\\&F=R(p^{2}+q^{2})-c^{2}=25(1+4)-25=100\end{aligned}}}
The equation of the parabola in Figure 2 is:
16
x
2
+
24
x
y
+
9
y
2
−
20
x
−
140
y
+
100
=
0.
{\displaystyle 16x^{2}+24xy+9y^{2}-20x-140y+100=0.}
Equation of directrix in normal form:
3
5
x
−
4
5
y
−
1
=
0.
{\displaystyle {\frac {3}{5}}x-{\frac {4}{5}}y-1=0.}
Distance from
f
o
c
u
s
{\displaystyle focus}
to
d
i
r
e
c
t
r
i
x
=
3
5
(
1
)
−
4
5
(
2
)
−
1
=
−
2.
{\displaystyle directrix={\frac {3}{5}}(1)-{\frac {4}{5}}(2)-1=-2.}
Distance from vertex to focus
=
1
=
1
4
K
{\displaystyle =1={\frac {1}{4K}}}
.
Therefore, curve has shape of
y
=
K
x
2
{\displaystyle y=Kx^{2}}
where
K
=
1
4
{\displaystyle K={\frac {1}{4}}}
.
Caution: An interesting situation occurs if the focus is on the directrix. Consider the directrix:
4
x
−
3
y
+
15
=
0
{\displaystyle 4x-3y+15=0}
and the focus
(
3
,
9
)
{\displaystyle (3,9)}
which is on the directrix.
a
=
4
,
b
=
−
3
,
c
=
15
,
p
=
3
,
q
=
9
{\displaystyle a=4,\ b=-3,\ c=15,\ p=3,\ q=9}
In this case the "parabola" has equation:
9
x
x
+
24
x
y
+
16
y
y
−
270
x
−
360
y
+
2025
=
0
{\displaystyle 9xx+24xy+16yy-270x-360y+2025=0}
.
This seems to be the equation of a parabola because
B
2
−
4
A
C
=
0
{\displaystyle B^{2}-4AC=0}
, but look closely.
9
x
x
+
24
x
y
+
16
y
y
−
270
x
−
360
y
+
2025
=
(
3
x
+
4
y
−
45
)
2
{\displaystyle 9xx+24xy+16yy-270x-360y+2025=(3x+4y-45)^{2}}
.
The result is a line through the focus and normal to the directrix.
If you solve for
p
,
q
,
c
{\displaystyle p,q,c}
using the algebraic solutions, you will produce the values
3
,
9
,
15
{\displaystyle 3,9,15}
as above.
However, the distance between focus and directrix =
4
5
x
+
−
3
5
y
+
3
{\displaystyle {\frac {4}{5}}x+{\frac {-3}{5}}y+3}
where
x
=
p
;
y
=
q
;
{\displaystyle x=p;\ y=q;}
distance
=
4
5
(
3
)
+
−
3
5
(
9
)
+
3
=
0.
{\displaystyle ={\frac {4}{5}}(3)+{\frac {-3}{5}}(9)+3=0.}
Reverse-Engineering the Parabola
edit
Figure 3: Parabola with 2 tangents parallel to axes. Tangent
A
B
C
:
y
=
16
3
{\displaystyle ABC:y={\frac {16}{3}}}
. Tangent
A
D
E
:
x
=
−
0.25
{\displaystyle ADE:x=-0.25}
. Point
A
{\displaystyle A}
on directrix, oblique, thin, black line.
{\displaystyle }
Line
A
F
G
{\displaystyle AFG}
perpendicular to focal chord
B
F
D
.
{\displaystyle BFD.}
Focus at
F
:
(
1
,
7
)
{\displaystyle F:(1,7)}
.
Given a parabola in form
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}
the aim is to produce the directrix and the focus.
We will solve the example shown in Figure 3:
9
x
2
−
24
x
y
+
16
y
2
+
70
x
−
260
y
+
1025
=
0
{\displaystyle 9x^{2}-24xy+16y^{2}+70x-260y+1025=0}
,
where:
A
=
9
B
=
−
24
C
=
16
D
=
70
E
=
−
260
F
=
1025
{\displaystyle {\begin{aligned}A&=9\\B&=-24\\C&=16\\D&=70\\E&=-260\\F&=1025\end{aligned}}}
a
=
C
=
4
;
b
=
−
B
2
a
=
−
(
−
24
)
8
=
3
{\displaystyle a={\sqrt {C}}=4;\ b={\frac {-B}{2a}}={\frac {-(-24)}{8}}=3}
.
Method 1. By analytical geometry
edit
Find two tangents that intersect at a right angle. The simplest to find are those that are parallel to the axes.
Method 2. By algebra
edit
A
=
b
2
B
=
−
2
a
b
C
=
a
2
D
=
−
2
(
a
c
+
p
R
)
E
=
−
2
(
b
c
+
q
R
)
F
=
R
(
p
2
+
q
2
)
−
c
2
B
2
−
4
A
C
=
0
R
=
a
2
+
b
2
{\displaystyle {\begin{aligned}A&=b^{2}\\B&=-2ab\\C&=a^{2}\\D&=-2(ac+pR)\\E&=-2(bc+qR)\\F&=R(p^{2}+q^{2})-c^{2}\\\\B&^{2}-4AC=0\\R&=a^{2}+b^{2}\end{aligned}}}
After rearranging the above values, there are three equations to be solved for three unknowns:
p
,
q
,
c
{\displaystyle p,q,c}
:
D
+
2
a
c
+
2
p
R
=
0
E
+
2
b
c
+
2
q
R
=
0
F
−
R
p
p
−
R
q
q
+
c
c
=
0
{\displaystyle {\begin{aligned}D+2ac+2pR=0\\E+2bc+2qR=0\\F-Rpp-Rqq+cc=0\end{aligned}}}
The solutions are:
c
=
4
F
R
−
(
D
D
+
E
E
)
4
D
a
+
4
E
b
{\displaystyle c={\frac {4FR-(DD+EE)}{4Da+4Eb}}}
p
=
−
(
D
+
2
a
c
)
2
R
{\displaystyle p={\frac {-(D+2ac)}{2R}}}
q
=
−
(
E
+
2
b
c
)
2
R
{\displaystyle q={\frac {-(E+2bc)}{2R}}}
Slope of the Parabola
edit
Consider parabola
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}
and line
y
=
m
x
+
c
.
{\displaystyle y=mx+c.}
Let point
(
X
,
Y
)
{\displaystyle (X,Y)}
be any point on the line.
Therefore
Y
=
m
X
+
c
;
c
=
Y
−
m
X
;
y
=
m
x
+
(
Y
−
m
X
)
.
{\displaystyle Y=mX+c;\ c=Y-mX;\ y=mx+(Y-mX).}
Let the line intersect the parabola.
Substitute the above value of
y
{\displaystyle y}
into the equation of the parabola and expand:
(
+
A
+
B
m
+
C
m
m
)
x
x
+
(
−
B
X
m
+
B
Y
−
2
C
X
m
m
+
2
C
Y
m
+
D
+
E
m
)
x
+
(
+
C
X
X
m
m
−
2
C
X
Y
m
+
C
Y
Y
−
E
X
m
+
E
Y
+
F
)
=
0
{\displaystyle {\begin{aligned}(+A+Bm+Cmm)&xx+\\(-BXm+BY-2CXmm+2CYm+D+Em)&x+\\(+CXXmm-2CXYm+CYY-EXm+EY+F)&=0\end{aligned}}}
We want the line to be tangent to the curve. Therefore
x
{\displaystyle x}
must have exactly one value and the discriminant is
0.
{\displaystyle 0.}
Discriminant =
(
−
B
X
m
+
B
Y
−
2
C
X
m
m
+
2
C
Y
m
+
D
+
E
m
)
(
−
B
X
m
+
B
Y
−
2
C
X
m
m
+
2
C
Y
m
+
D
+
E
m
)
−
4
(
+
A
+
B
m
+
C
m
m
)
(
+
C
X
X
m
m
−
2
C
X
Y
m
+
C
Y
Y
−
E
X
m
+
E
Y
+
F
)
=
0
{\displaystyle {\begin{aligned}(-BXm+BY-2CXmm+2CYm+D+Em)(-BXm+BY-2CXmm+2CYm+D+Em)&\\-4(+A+Bm+Cmm)(+CXXmm-2CXYm+CYY-EXm+EY+F)&=0\end{aligned}}}
The above discriminant is a quadratic in
m
{\displaystyle m}
:
(
−
4
A
C
X
X
+
B
B
X
X
+
2
B
E
X
−
4
C
D
X
−
4
C
F
+
E
E
)
m
m
+
(
+
8
A
C
X
Y
+
4
A
E
X
−
2
B
B
X
Y
−
2
B
D
X
−
2
B
E
Y
−
4
B
F
+
4
C
D
Y
+
2
D
E
)
m
+
(
−
4
A
C
Y
Y
−
4
A
E
Y
−
4
A
F
+
B
B
Y
Y
+
2
B
D
Y
+
D
D
)
=
0
{\displaystyle {\begin{aligned}(-4ACXX+BBXX+2BEX-4CDX-4CF+EE)&mm+\\(+8ACXY+4AEX-2BBXY-2BDX-2BEY-4BF+4CDY+2DE)&m+\\(-4ACYY-4AEY-4AF+BBYY+2BDY+DD)&=0\end{aligned}}}
m
=
−
(
+
8
A
C
X
Y
+
4
A
E
X
−
2
B
B
X
Y
−
2
B
D
X
−
2
B
E
Y
−
4
B
F
+
4
C
D
Y
+
2
D
E
)
±
R
2
(
−
4
A
C
X
X
+
B
B
X
X
+
2
B
E
X
−
4
C
D
X
−
4
C
F
+
E
E
)
{\displaystyle m={\frac {-(+8ACXY+4AEX-2BBXY-2BDX-2BEY-4BF+4CDY+2DE)\pm {\sqrt {R}}}{2(-4ACXX+BBXX+2BEX-4CDX-4CF+EE)}}}
where:
R
=
16
(
A
X
X
+
B
X
Y
+
C
Y
Y
+
D
X
+
E
Y
+
F
)
(
−
4
A
C
F
+
A
E
E
+
B
B
F
−
B
D
E
+
C
D
D
)
{\displaystyle {\begin{aligned}R=&\ 16(AXX+BXY+CYY+DX+EY+F)(-4ACF+AEE+BBF-BDE+CDD)\end{aligned}}}
If the point
(
X
,
Y
)
{\displaystyle (X,Y)}
is on the curve, then the line touches the curve at
(
X
,
Y
)
{\displaystyle (X,Y)}
and:
(
A
X
X
+
B
X
Y
+
C
Y
Y
+
D
X
+
E
Y
+
F
)
=
0
;
R
=
0
m
=
−
(
+
8
A
C
X
Y
+
4
A
E
X
−
2
B
B
X
Y
−
2
B
D
X
−
2
B
E
Y
−
4
B
F
+
4
C
D
Y
+
2
D
E
)
2
(
−
4
A
C
X
X
+
B
B
X
X
+
2
B
E
X
−
4
C
D
X
−
4
C
F
+
E
E
)
=
2
B
B
X
Y
−
8
A
C
X
Y
+
2
B
D
X
−
4
A
E
X
+
2
B
E
Y
−
4
C
D
Y
+
4
B
F
−
2
D
E
2
(
B
B
X
X
−
4
A
C
X
X
+
2
B
E
X
−
4
C
D
X
+
E
E
−
4
C
F
)
=
(
B
B
−
4
A
C
)
X
Y
+
(
B
D
−
2
A
E
)
X
+
(
B
E
−
2
C
D
)
Y
+
2
B
F
−
D
E
(
B
B
−
4
A
C
)
X
X
+
(
2
B
E
−
4
C
D
)
X
+
E
E
−
4
C
F
=
(
B
D
−
2
A
E
)
X
+
(
B
E
−
2
C
D
)
Y
+
2
B
F
−
D
E
(
2
B
E
−
4
C
D
)
X
+
E
E
−
4
C
F
{\displaystyle {\begin{aligned}(AX&X+BXY+CYY+DX+EY+F)=0;\ R=0\\\\m=&\ {\frac {-(+8ACXY+4AEX-2BBXY-2BDX-2BEY-4BF+4CDY+2DE)}{2(-4ACXX+BBXX+2BEX-4CDX-4CF+EE)}}\\\\=&\ {\frac {2BBXY-8ACXY+2BDX-4AEX+2BEY-4CDY+4BF-2DE}{2(BBXX-4ACXX+2BEX-4CDX+EE-4CF)}}\\\\=&\ {\frac {(BB-4AC)XY+(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(BB-4AC)XX+(2BE-4CD)X+EE-4CF}}\\\\=&\ {\frac {(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(2BE-4CD)X+EE-4CF}}\end{aligned}}}
because
B
2
−
4
A
C
=
0
{\displaystyle B^{2}-4AC=0}
for a parabola.
When slope is displayed in this format, we see that slope is vertical if
(
2
B
E
−
4
C
D
)
X
+
E
E
−
4
C
F
=
0.
{\displaystyle (2BE-4CD)X+EE-4CF=0.}
The line
x
=
4
C
F
−
E
E
2
B
E
−
4
C
D
{\displaystyle x={\frac {4CF-EE}{2BE-4CD}}}
is tangent to the curve.
Let the equation of a line be:
x
=
M
y
+
c
{\displaystyle x=My+c}
in which case
M
=
1
m
.
{\displaystyle M={\frac {1}{m}}.}
By using calculations similar to the above it can be shown that:
M
=
(
B
D
−
2
A
E
)
X
+
(
B
E
−
2
C
D
)
Y
+
2
B
F
−
D
E
(
2
B
D
−
4
A
E
)
Y
+
D
D
−
4
A
F
{\displaystyle M={\frac {(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(2BD-4AE)Y+DD-4AF}}}
.
m
=
1
M
{\displaystyle m={\frac {1}{M}}}
therefore:
m
=
(
2
B
D
−
4
A
E
)
Y
+
D
D
−
4
A
F
(
B
D
−
2
A
E
)
X
+
(
B
E
−
2
C
D
)
Y
+
2
B
F
−
D
E
{\displaystyle m={\frac {(2BD-4AE)Y+DD-4AF}{(BD-2AE)X+(BE-2CD)Y+2BF-DE}}}
When slope is displayed in this format, we see that slope is zero if
(
2
B
D
−
4
A
E
)
Y
+
D
D
−
4
A
F
=
0
{\displaystyle (2BD-4AE)Y+DD-4AF=0}
.
The line
y
=
4
A
F
−
D
D
2
B
D
−
4
A
E
{\displaystyle y={\frac {4AF-DD}{2BD-4AE}}}
is tangent to the curve.
This examination of the parabola has produced two expressions for slope of the parabola:
m
=
(
2
B
D
−
4
A
E
)
y
+
D
D
−
4
A
F
(
B
D
−
2
A
E
)
x
+
(
B
E
−
2
C
D
)
y
+
2
B
F
−
D
E
=
(
B
D
−
2
A
E
)
x
+
(
B
E
−
2
C
D
)
y
+
2
B
F
−
D
E
(
2
B
E
−
4
C
D
)
x
+
E
E
−
4
C
F
{\displaystyle m={\frac {(2BD-4AE)y+DD-4AF}{(BD-2AE)x+(BE-2CD)y+2BF-DE}}={\frac {(BD-2AE)x+(BE-2CD)y+2BF-DE}{(2BE-4CD)x+EE-4CF}}}
.
where the point
(
x
,
y
)
{\displaystyle (x,y)}
is any arbitrary point on the curve.
Therefore
m
=
±
(
2
B
D
−
4
A
E
)
y
+
D
D
−
4
A
F
(
2
B
E
−
4
C
D
)
x
+
E
E
−
4
C
F
{\displaystyle m=\pm {\sqrt {\frac {(2BD-4AE)y+DD-4AF}{(2BE-4CD)x+EE-4CF}}}}
. This formula for
m
{\displaystyle m}
contains both tangents parallel to the axes and is derived without calculus.
The formula from calculus below is simpler and unambiguous concerning sign.
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
d
d
x
(
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
)
=
0
A
(
2
x
)
+
B
(
x
d
y
d
x
+
y
d
x
d
x
)
+
C
(
2
y
d
y
d
x
)
+
D
+
E
(
d
y
d
x
)
=
0
A
(
2
x
)
+
B
(
x
d
y
d
x
+
y
)
+
C
(
2
y
d
y
d
x
)
+
D
+
E
(
d
y
d
x
)
=
0
2
A
x
+
B
x
d
y
d
x
+
B
y
+
2
C
y
d
y
d
x
+
D
+
E
d
y
d
x
=
0
d
y
d
x
(
B
x
+
2
C
y
+
E
)
=
−
(
2
A
x
+
B
y
+
D
)
m
=
d
y
d
x
=
−
(
2
A
x
+
B
y
+
D
)
B
x
+
2
C
y
+
E
{\displaystyle {\begin{aligned}Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\\\\{\frac {d}{dx}}(Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F)=0\\\\A(2x)+B(x{\frac {dy}{dx}}+y{\frac {dx}{dx}})+C(2y{\frac {dy}{dx}})+D+E({\frac {dy}{dx}})=0\\\\A(2x)+B(x{\frac {dy}{dx}}+y)+C(2y{\frac {dy}{dx}})+D+E({\frac {dy}{dx}})=0\\\\2Ax+Bx{\frac {dy}{dx}}+By+2Cy{\frac {dy}{dx}}+D+E{\frac {dy}{dx}}=0\\\\{\frac {dy}{dx}}(Bx+2Cy+E)=-(2Ax+By+D)\\\\m={\frac {dy}{dx}}={\frac {-(2Ax+By+D)}{Bx+2Cy+E}}\end{aligned}}}
The slope of the parabola is
0
{\displaystyle 0}
where
(
2
B
D
−
4
A
E
)
y
+
D
D
−
4
A
F
=
0
;
2
A
x
+
B
y
+
D
=
0
;
{\displaystyle (2BD-4AE)y+DD-4AF=0;\ 2Ax+By+D=0;}
or:
y
=
4
A
F
−
D
D
2
B
D
−
4
A
E
;
x
=
−
(
B
y
+
D
)
2
A
.
{\displaystyle y={\frac {4AF-DD}{2BD-4AE}};\ x={\frac {-(By+D)}{2A}}.}
The slope of the parabola is vertical where
(
2
B
E
−
4
C
D
)
x
+
E
E
−
4
C
F
=
0
;
B
x
+
2
C
y
+
E
=
0
;
{\displaystyle (2BE-4CD)x+EE-4CF=0;\ Bx+2Cy+E=0;}
or:
x
=
4
C
F
−
E
E
2
B
E
−
4
C
D
;
y
=
−
(
B
x
+
E
)
2
C
.
{\displaystyle x={\frac {4CF-EE}{2BE-4CD}};\ y={\frac {-(Bx+E)}{2C}}.}
Caution:
If the curve is
y
=
K
x
2
{\displaystyle y=Kx^{2}}
, the slope can never be vertical.
If the curve is
x
=
K
y
2
{\displaystyle x=Ky^{2}}
, the slope can never be
0
{\displaystyle 0}
.
If
B
=
C
=
0
{\displaystyle B=C=0}
and
E
=
−
1
{\displaystyle E=-1}
, the equation of the parabola becomes:
y
=
A
x
2
+
D
x
+
F
{\displaystyle y=Ax^{2}+Dx+F}
and:
m
=
(
2
B
D
−
4
A
E
)
y
+
D
D
−
4
A
F
(
B
D
−
2
A
E
)
x
+
(
B
E
−
2
C
D
)
y
+
2
B
F
−
D
E
=
(
−
4
A
(
−
1
)
)
y
+
D
D
−
4
A
F
(
−
2
A
(
−
1
)
)
x
−
D
(
−
1
)
=
4
A
y
+
D
D
−
4
A
F
2
A
x
+
D
=
4
A
(
A
x
2
+
D
x
+
F
)
+
D
D
−
4
A
F
2
A
x
+
D
=
4
A
2
x
2
+
4
A
D
x
+
4
A
F
+
D
D
−
4
A
F
2
A
x
+
D
=
4
A
2
x
2
+
4
A
D
x
+
D
D
2
A
x
+
D
=
2
A
x
+
D
m
=
(
B
D
−
2
A
E
)
x
+
(
B
E
−
2
C
D
)
y
+
2
B
F
−
D
E
(
2
B
E
−
4
C
D
)
x
+
E
E
−
4
C
F
=
(
(
0
)
D
−
2
A
(
−
1
)
)
x
+
(
(
0
)
E
−
2
(
0
)
D
)
y
+
2
(
0
)
F
−
D
(
−
1
)
(
2
(
0
)
E
−
4
(
0
)
D
)
x
+
(
−
1
)
(
−
1
)
−
4
(
0
)
F
=
(
−
2
A
(
−
1
)
)
x
−
D
(
−
1
)
(
−
1
)
(
−
1
)
=
2
A
x
+
D
m
=
−
(
2
A
x
+
B
y
+
D
)
B
x
+
2
C
y
+
E
=
−
(
2
A
x
+
(
0
)
y
+
D
)
(
0
)
x
+
2
(
0
)
y
+
(
−
1
)
=
2
A
x
+
D
{\displaystyle {\begin{aligned}m=&\ {\frac {(2BD-4AE)y+DD-4AF}{(BD-2AE)x+(BE-2CD)y+2BF-DE}}\\\\=&\ {\frac {(-4A(-1))y+DD-4AF}{(-2A(-1))x-D(-1)}}={\frac {4Ay+DD-4AF}{2Ax+D}}\\\\=&\ {\frac {4A(Ax^{2}+Dx+F)+DD-4AF}{2Ax+D}}={\frac {4A^{2}x^{2}+4ADx+4AF+DD-4AF}{2Ax+D}}\\\\=&\ {\frac {4A^{2}x^{2}+4ADx+DD}{2Ax+D}}\\\\=&\ 2Ax+D\\\\m=&\ {\frac {(BD-2AE)x+(BE-2CD)y+2BF-DE}{(2BE-4CD)x+EE-4CF}}\\\\=&\ {\frac {((0)D-2A(-1))x+((0)E-2(0)D)y+2(0)F-D(-1)}{(2(0)E-4(0)D)x+(-1)(-1)-4(0)F}}\\\\=&\ {\frac {(-2A(-1))x-D(-1)}{(-1)(-1)}}\\\\=&\ 2Ax+D\\\\m=&\ {\frac {-(2Ax+By+D)}{Bx+2Cy+E}}\\\\=&\ {\frac {-(2Ax+(0)y+D)}{(0)x+2(0)y+(-1)}}\\\\=&\ 2Ax+D\end{aligned}}}
Point at given slope
edit
Given parabola defined by
A
,
B
,
C
,
D
,
E
,
F
{\displaystyle A,B,C,D,E,F}
and slope
m
=
y
v
a
l
u
e
x
v
a
l
u
e
{\displaystyle m={\frac {yvalue}{xvalue}}}
where at least one of
y
v
a
l
u
e
,
x
v
a
l
u
e
{\displaystyle yvalue,xvalue}
is non-zero, calculate point at which the slope is
m
.
{\displaystyle m.}
Let
m
=
−
(
2
A
x
+
B
y
+
D
)
B
x
+
2
C
y
+
E
=
s
t
=
y
v
a
l
u
e
x
v
a
l
u
e
{\displaystyle m={\frac {-(2Ax+By+D)}{Bx+2Cy+E}}={\frac {s}{t}}={\frac {yvalue}{xvalue}}}
Then
s
(
B
x
+
2
C
y
+
E
)
+
t
(
2
A
x
+
B
y
+
D
)
=
0.
{\displaystyle s(Bx+2Cy+E)+t(2Ax+By+D)=0.}
Let
G
=
(
B
s
+
2
A
t
)
;
H
=
(
2
C
s
+
B
t
)
;
I
=
(
E
s
+
D
t
)
.
{\displaystyle G=(Bs+2At);\ H=(2Cs+Bt);\ I=(Es+Dt).}
Then
G
x
+
H
y
+
I
=
0
…
(
1
)
{\displaystyle Gx+Hy+I=0\ \dots \ (1)}
Substitute in the equation of the parabola and
y
=
−
(
A
I
I
−
D
G
I
+
F
G
G
)
(
2
A
H
I
−
B
G
I
−
D
G
H
+
E
G
G
)
…
(
2
)
{\displaystyle y={\frac {-(AII-DGI+FGG)}{(2AHI-BGI-DGH+EGG)}}\ \dots \ (2)}
As shown below, with a little manipulation of the data, the same formula can be used to calculate
x
.
{\displaystyle x.}
# python code
def pointAtGivenSlope ( parabola , tangent ) :
s , t = tangent
if s == t == 0 :
print ( 'pointAtGivenSlope(): both s,t can not be 0.' )
return None
def calculate_y ( parabola , tangent ) :
A , B , C , D , E , F = parabola
s , t = tangent
G = B * s + 2 * A * t ; H = 2 * C * s + B * t ; I = E * s + D * t
return - ( A * I * I - D * G * I + F * G * G ) / ( 2 * A * H * I - B * G * I - D * G * H + E * G * G )
y = calculate_y ( parabola , tangent )
A , B , C , D , E , F = parabola
x = calculate_y (( C , B , A , E , D , F ), ( t , s ))
return x , y
A parabola is defined as
9
x
2
−
24
x
y
+
16
y
2
+
70
x
−
260
y
+
1025
=
0.
{\displaystyle 9x^{2}-24xy+16y^{2}+70x-260y+1025=0.}
Calculate coordinates of vertex.
At vertex tangent has same slope as directrix.
# python code
parabola = A , B , C , D , E , F = 9 , - 24 , 16 , 70 , - 260 , 1025
a = C ** .5
b = - B / ( 2 * a )
tangent = - a , b
result = pointAtGivenSlope ( parabola , tangent )
print ( result )
Calculate point at which tangent is vertical.
# python code
parabola = A , B , C , D , E , F = 9 , - 24 , 16 , 70 , - 260 , 1025
tangent = 1 , 0
result = pointAtGivenSlope ( parabola , tangent )
print ( result )
Parabola and any chord
edit
Figure 4: Parabola and any chord. Origin
(
0
,
0
)
{\displaystyle (0,0)}
at point
O
;
{\displaystyle O;}
curve
:
{\displaystyle :}
y
=
K
x
2
;
{\displaystyle y=Kx^{2};}
chord
I
J
{\displaystyle IJ}
:
y
=
m
x
+
c
;
{\displaystyle :\ y=mx+c;}
point
L
{\displaystyle L}
:
(
0
,
c
)
;
{\displaystyle :\ (0,c);}
2 tangents
:
{\displaystyle :}
I
N
,
J
N
;
{\displaystyle IN,JN;}
point
N
{\displaystyle N}
:
(
m
2
K
,
−
c
)
{\displaystyle :\ ({\frac {m}{2K}},-c)}
Refer to Figure 4.
The curve has equation:
y
=
K
x
2
.
{\displaystyle y=Kx^{2}.}
The chord
I
J
{\displaystyle IJ}
has equation:
y
=
m
x
+
c
{\displaystyle y=mx+c}
.
Point
L
{\displaystyle L}
has coordinates
(
0
,
c
)
.
{\displaystyle (0,c).}
Line
I
N
{\displaystyle IN}
is tangent to the curve at
I
.
{\displaystyle I.}
Line
J
N
{\displaystyle JN}
is tangent to the curve at
J
.
{\displaystyle J.}
This section shows that point
N
{\displaystyle N}
has coordinates
(
m
2
K
,
−
c
)
.
{\displaystyle ({\frac {m}{2K}},-c).}
y
=
K
x
2
=
m
x
+
c
K
x
2
−
m
x
−
c
=
0
x
=
m
±
m
2
+
4
K
c
2
K
=
m
±
R
2
K
{\displaystyle {\begin{aligned}y=&\ Kx^{2}=mx+c\\\\Kx^{2}&-mx-c=0\\\\x=&\ {\frac {m\pm {\sqrt {m^{2}+4Kc}}}{2K}}\\\\=&\ {\frac {m\pm R}{2K}}\end{aligned}}}
where
R
=
m
2
+
4
K
c
{\displaystyle R={\sqrt {m^{2}+4Kc}}}
Point
I
{\displaystyle I}
has coordinates
(
x
1
,
y
1
)
{\displaystyle (x_{1},y_{1})}
where:
x
1
=
m
−
R
2
K
{\displaystyle x_{1}={\frac {m-R}{2K}}}
,
y
1
=
K
x
1
2
=
K
(
m
−
R
2
K
)
(
m
−
R
2
K
)
=
m
2
−
2
m
R
+
R
2
4
K
=
m
2
−
m
R
+
2
K
c
2
K
{\displaystyle y_{1}=Kx_{1}^{2}=K({\frac {m-R}{2K}})({\frac {m-R}{2K}})={\frac {m^{2}-2mR+R^{2}}{4K}}={\frac {m^{2}-mR+2Kc}{2K}}}
,
and slope of tangent
I
N
=
s
1
=
2
K
x
1
=
m
−
R
.
{\displaystyle IN=s_{1}=2Kx_{1}=m-R.}
Point
J
{\displaystyle J}
has coordinates
(
x
2
,
y
2
)
{\displaystyle (x_{2},y_{2})}
where:
x
2
=
m
+
R
2
K
{\displaystyle x_{2}={\frac {m+R}{2K}}}
,
y
2
=
m
2
+
m
R
+
2
K
c
2
K
{\displaystyle y_{2}={\frac {m^{2}+mR+2Kc}{2K}}}
,
and slope of tangent
J
N
=
s
2
=
m
+
R
.
{\displaystyle JN=s_{2}=m+R.}
Points
D
,
E
{\displaystyle D,E}
have coordinates
(
x
1
,
0
)
,
(
x
2
,
0
)
.
{\displaystyle (x_{1},0),(x_{2},0).}
Equation of tangent
I
N
:
{\displaystyle IN:}
y
=
s
1
x
+
c
1
c
1
=
y
1
−
s
1
x
1
=
m
2
−
m
R
+
2
K
c
2
K
−
(
m
−
R
)
(
m
−
R
2
K
)
=
−
(
m
2
−
m
R
+
2
K
c
)
2
K
y
=
(
m
−
R
)
x
−
m
2
−
m
R
+
2
K
c
2
K
{\displaystyle {\begin{aligned}y=&\ s_{1}x+c_{1}\\c_{1}=&\ y_{1}-s_{1}x_{1}={\frac {m^{2}-mR+2Kc}{2K}}-(m-R)({\frac {m-R}{2K}})={\frac {-(m^{2}-mR+2Kc)}{2K}}\\y=&\ (m-R)x-{\frac {m^{2}-mR+2Kc}{2K}}\end{aligned}}}
Equation of tangent
J
N
:
{\displaystyle JN:}
y
=
(
m
+
R
)
x
−
m
2
+
m
R
+
2
K
c
2
K
{\displaystyle {\begin{aligned}y=(m+R)x-{\frac {m^{2}+mR+2Kc}{2K}}\end{aligned}}}
At point of intersection
N
,
(
m
+
R
)
x
−
m
2
+
m
R
+
2
K
c
2
K
=
(
m
−
R
)
x
−
m
2
−
m
R
+
2
K
c
2
K
;
x
=
m
2
K
.
{\displaystyle N,\ (m+R)x-{\frac {m^{2}+mR+2Kc}{2K}}=(m-R)x-{\frac {m^{2}-mR+2Kc}{2K}};\ x={\frac {m}{2K}}.}
Review the
X
{\displaystyle X}
coordinates of points
D
,
E
{\displaystyle D,E}
:
(
m
−
R
2
K
,
m
+
R
2
K
)
{\displaystyle :\ ({\frac {m-R}{2K}},{\frac {m+R}{2K}})}
.
The line
N
G
H
{\displaystyle NGH}
with equation
x
=
m
2
K
{\displaystyle x={\frac {m}{2K}}}
bisects the line segment
D
E
{\displaystyle DE}
and also the chord
I
J
{\displaystyle IJ}
at point
H
{\displaystyle H}
.
Any chord parallel to
I
J
{\displaystyle IJ}
has two tangents that intersect on the line
x
=
m
2
K
{\displaystyle x={\frac {m}{2K}}}
.
Any chord parallel to
I
J
{\displaystyle IJ}
is bisected by the line
x
=
m
2
K
{\displaystyle x={\frac {m}{2K}}}
.
The
Y
{\displaystyle Y}
coordinate of point
N
:
{\displaystyle N:}
{\displaystyle }
y
=
s
1
(
m
2
K
)
+
c
1
=
(
m
−
R
)
(
m
2
K
)
−
m
2
−
m
R
+
2
K
c
2
K
=
m
2
−
m
R
2
K
−
m
2
−
m
R
+
2
K
c
2
K
=
−
2
K
c
2
K
=
−
c
{\displaystyle {\begin{aligned}y=&\ s_{1}({\frac {m}{2K}})+c_{1}\\=&\ (m-R)({\frac {m}{2K}})-{\frac {m^{2}-mR+2Kc}{2K}}\\=&\ {\frac {m^{2}-mR}{2K}}-{\frac {m^{2}-mR+2Kc}{2K}}\\=&\ {\frac {-2Kc}{2K}}\\=&\ -c\end{aligned}}}
Any chord that passes through the point
L
(
0
,
c
)
{\displaystyle L\ (0,c)}
has two tangents that intersect on the line
y
=
−
c
{\displaystyle y=-c}
.
Angle
I
N
J
{\displaystyle INJ}
Using:
tan
(
A
−
B
)
=
tan
(
A
)
−
tan
(
B
)
1
+
tan
(
A
)
tan
(
B
)
tan
(
∠
I
N
J
)
=
(
m
−
R
)
−
(
m
+
R
)
1
+
(
m
−
R
)
(
m
+
R
)
=
−
2
R
1
+
(
m
2
−
R
2
)
=
−
2
R
1
+
m
2
−
(
m
2
+
4
K
c
)
=
−
2
R
1
−
4
K
c
=
2
m
2
+
4
K
c
4
K
c
−
1
{\displaystyle {\begin{aligned}\tan(A-B)=&\ {\frac {\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}}\\\\\tan(\angle INJ)=&\ {\frac {(m-R)-(m+R)}{1+(m-R)(m+R)}}\\\\=&\ {\frac {-2R}{1+(m^{2}-R^{2})}}\\\\=&\ {\frac {-2R}{1+m^{2}-(m^{2}+4Kc)}}\\\\=&\ {\frac {-2R}{1-4Kc}}\\\\=&\ {\frac {2{\sqrt {m^{2}+4Kc}}}{4Kc-1}}\end{aligned}}}
If
4
K
c
>
1
{\displaystyle 4Kc>1}
, point
L
{\displaystyle L}
is above the
f
o
c
u
s
,
tan
(
∠
I
N
J
)
{\displaystyle focus,\ \tan(\angle INJ)}
is positive
and
0
{\displaystyle 0}
°
<
∠
I
N
J
<
90
{\displaystyle <\angle INJ<90}
°.
∠
I
N
J
{\displaystyle \angle INJ}
is acute and, as
m
{\displaystyle m}
increases,
∠
I
N
J
{\displaystyle \angle INJ}
increases, approaching
90
{\displaystyle 90}
°.
If
4
K
c
==
1
{\displaystyle 4Kc==1}
, point
L
{\displaystyle L}
is on the
f
o
c
u
s
,
tan
(
∠
I
N
J
)
=
2
m
2
+
1
0
,
∠
I
N
J
=
90
{\displaystyle focus,\ \tan(\angle INJ)={\frac {2{\sqrt {m^{2}+1}}}{0}},\ \angle INJ=90}
°
and the line
y
=
−
c
{\displaystyle y=-c}
is the
d
i
r
e
c
t
r
i
x
{\displaystyle directrix}
.
If
4
K
c
<
1
{\displaystyle 4Kc<1}
, point
L
{\displaystyle L}
is below the
f
o
c
u
s
,
tan
(
∠
I
N
J
)
{\displaystyle focus,\ \tan(\angle INJ)}
is negative
and
90
{\displaystyle 90}
°
<
∠
I
N
J
<
180
{\displaystyle <\angle INJ<180}
°.
∠
I
N
J
{\displaystyle \angle INJ}
is obtuse and, as
m
{\displaystyle m}
increases,
∠
I
N
J
{\displaystyle \angle INJ}
decreases, approaching
90
{\displaystyle 90}
°.
Parabola and two tangents
edit
Figure 5: Parabola and two tangents. Origin
(
0
,
0
)
{\displaystyle (0,0)}
at point
O
;
{\displaystyle O;}
curve
:
{\displaystyle :}
y
=
K
x
2
;
{\displaystyle y=Kx^{2};}
point
N
{\displaystyle N}
:
(
h
,
−
c
)
;
{\displaystyle :\ (h,-c);}
2 tangents
:
{\displaystyle :}
N
I
,
N
J
;
{\displaystyle NI,NJ;}
chord
I
J
{\displaystyle IJ}
:
y
=
2
K
h
x
+
c
;
{\displaystyle :\ y=2Khx+c;}
point
L
{\displaystyle L}
:
(
0
,
c
)
;
{\displaystyle :\ (0,c);}
Refer to Figure 5.
The curve has equation:
y
=
K
x
2
.
{\displaystyle y=Kx^{2}.}
Point
N
{\displaystyle N}
with coordinates
(
h
,
−
c
)
{\displaystyle (h,-c)}
is any point on the line
y
=
−
c
.
{\displaystyle y=-c.}
Line
N
I
{\displaystyle NI}
is tangent to the curve at point
I
(
x
1
,
y
1
)
{\displaystyle I\ (x_{1},y_{1})}
.
Line
N
J
{\displaystyle NJ}
is tangent to the curve at point
J
(
x
2
,
y
2
)
{\displaystyle J\ (x_{2},y_{2})}
.
This section shows that the chord
I
J
{\displaystyle IJ}
passes through the point
(
0
,
c
)
.
{\displaystyle (0,c).}
Equation of any line through point
N
:
y
=
m
x
−
c
−
m
h
{\displaystyle N:\ y=mx-c-mh}
Let this line intersect the curve:
y
=
K
x
2
=
m
x
−
c
−
m
h
{\displaystyle y=Kx^{2}=mx-c-mh}
K
x
2
−
m
x
+
c
+
m
h
=
0
{\displaystyle Kx^{2}-mx+c+mh=0}
x
=
m
±
m
2
−
4
K
(
m
h
+
c
)
2
K
=
m
±
m
2
−
4
K
m
h
−
4
K
c
2
K
{\displaystyle x={\frac {m\pm {\sqrt {m^{2}-4K(mh+c)}}}{2K}}={\frac {m\pm {\sqrt {m^{2}-4Kmh-4Kc}}}{2K}}}
We want this line to be a tangent to the curve, therefore
x
{\displaystyle x}
has exactly one value and the discriminant is
0
{\displaystyle 0}
:
m
2
−
4
K
m
h
−
4
K
c
=
0
m
=
4
K
h
±
(
4
K
h
)
2
−
4
(
1
)
(
−
4
K
c
)
2
=
4
K
h
±
16
K
2
h
2
+
16
K
c
2
=
2
K
h
±
2
R
{\displaystyle {\begin{aligned}m^{2}-&4Kmh-4Kc=0\\\\m=&\ {\frac {4Kh\pm {\sqrt {(4Kh)^{2}-4(1)(-4Kc)}}}{2}}\\\\=&\ {\frac {4Kh\pm {\sqrt {16K^{2}h^{2}+16Kc}}}{2}}\\\\=&\ 2Kh\pm 2R\end{aligned}}}
where
R
=
K
2
h
2
+
K
c
{\displaystyle R={\sqrt {K^{2}h^{2}+Kc}}}
m
1
=
2
K
h
−
2
R
=
{\displaystyle m_{1}=2Kh-2R=}
slope of tangent
N
I
{\displaystyle NI}
.
m
2
=
2
K
h
+
2
R
=
{\displaystyle m_{2}=2Kh+2R=}
slope of tangent
N
J
{\displaystyle NJ}
.
x
=
m
2
K
x
1
=
m
1
2
K
=
2
K
h
−
2
R
2
K
=
K
h
−
R
K
y
1
=
K
x
1
2
=
K
(
K
h
−
R
K
)
(
K
h
−
R
K
)
=
(
K
h
−
R
)
(
K
h
−
R
)
K
x
2
=
K
h
+
R
K
y
2
=
(
K
h
+
R
)
(
K
h
+
R
)
K
{\displaystyle {\begin{aligned}x=&\ {\frac {m}{2K}}\\\\x_{1}=&\ {\frac {m_{1}}{2K}}={\frac {2Kh-2R}{2K}}={\frac {Kh-R}{K}}\\\\y_{1}=&\ Kx_{1}^{2}=K({\frac {Kh-R}{K}})({\frac {Kh-R}{K}})={\frac {(Kh-R)(Kh-R)}{K}}\\\\x_{2}=&\ {\frac {Kh+R}{K}}\\\\y_{2}=&\ {\frac {(Kh+R)(Kh+R)}{K}}\end{aligned}}}
Slope of chord
I
J
{\displaystyle IJ}
=
y
2
−
y
1
x
2
−
x
1
=
(
(
K
h
+
R
)
(
K
h
+
R
)
K
−
(
K
h
−
R
)
(
K
h
−
R
)
K
)
/
(
K
h
+
R
K
−
K
h
−
R
K
)
=
(
K
K
h
h
+
2
K
h
R
+
R
R
−
(
K
K
h
h
−
2
K
h
R
+
R
R
)
K
)
/
(
2
R
K
)
=
(
4
K
h
R
K
)
(
K
2
R
)
=
2
K
h
{\displaystyle {\begin{aligned}=&\ {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\\\\=&\ ({\frac {(Kh+R)(Kh+R)}{K}}-{\frac {(Kh-R)(Kh-R)}{K}})/({\frac {Kh+R}{K}}-{\frac {Kh-R}{K}})\\\\=&\ ({\frac {KKhh+2KhR+RR-(KKhh-2KhR+RR)}{K}})/({\frac {2R}{K}})\\\\=&\ ({\frac {4KhR}{K}})({\frac {K}{2R}})\\\\=&\ 2Kh\end{aligned}}}
Intercept of chord
I
J
{\displaystyle IJ}
on the
Y
{\displaystyle Y}
axis
=
y
1
−
2
K
h
x
1
=
(
K
h
−
R
)
(
K
h
−
R
)
K
−
2
K
h
K
h
−
R
K
=
K
K
h
h
−
2
K
h
R
+
R
R
K
−
2
K
h
(
K
h
−
R
)
K
=
K
K
h
h
−
2
K
h
R
+
R
R
−
2
K
h
K
h
+
2
K
h
R
K
=
K
K
h
h
+
R
R
−
2
K
h
K
h
K
=
−
K
K
h
h
+
K
K
h
h
+
K
c
K
=
+
K
c
K
=
c
{\displaystyle {\begin{aligned}=&\ y_{1}-2Khx_{1}\\\\=&\ {\frac {(Kh-R)(Kh-R)}{K}}-2Kh{\frac {Kh-R}{K}}\\\\=&\ {\frac {KKhh-2KhR+RR}{K}}-{\frac {2Kh(Kh-R)}{K}}\\\\=&\ {\frac {KKhh-2KhR+RR-2KhKh+2KhR}{K}}\\\\=&\ {\frac {KKhh+RR-2KhKh}{K}}\\\\=&\ {\frac {-KKhh+KKhh+Kc}{K}}\\\\=&\ {\frac {+Kc}{K}}\\\\=&\ c\end{aligned}}}
Angle
I
N
J
{\displaystyle INJ}
If
∠
I
N
J
==
90
{\displaystyle \angle INJ==90}
°
(
m
1
)
(
m
2
)
=
−
1
(
2
K
h
−
2
R
)
(
2
K
h
+
2
R
)
=
−
1
4
(
K
K
h
h
−
R
R
)
=
−
1
4
(
K
K
h
h
−
(
K
K
h
h
+
K
c
)
)
=
−
1
4
(
−
K
c
)
=
−
1
4
K
c
=
1
{\displaystyle {\begin{aligned}&(m_{1})(m_{2})=-1\\\\&(2Kh-2R)(2Kh+2R)=-1\\\\&4(KKhh-RR)=-1\\\\&4(KKhh-(KKhh+Kc))=-1\\\\&4(-Kc)=-1\\\\&4Kc=1\end{aligned}}}
In the basic parabola
y
=
x
2
4
q
{\displaystyle y={\frac {x^{2}}{4q}}}
where the
f
o
c
u
s
{\displaystyle focus}
has coordinates
(
0
,
q
)
{\displaystyle (0,q)}
.
y
=
K
x
2
∴
K
=
1
4
q
{\displaystyle y=Kx^{2}\therefore K={\frac {1}{4q}}}
or
4
K
q
=
1.
{\displaystyle 4Kq=1.}
If
4
K
c
==
1
,
{\displaystyle 4Kc==1,}
then
(
0
,
c
)
{\displaystyle (0,c)}
and
(
0
,
q
)
{\displaystyle (0,q)}
are the same point, the chord
I
J
{\displaystyle IJ}
passes through the
f
o
c
u
s
{\displaystyle focus}
and the line
y
=
−
c
{\displaystyle y=-c}
is the
d
i
r
e
c
t
r
i
x
{\displaystyle directrix}
.
Area enclosed between parabola and chord
edit
Figure 6: The Parabola:
y
=
x
2
{\displaystyle y=x^{2}}
Chord
D
C
{\displaystyle DC}
, parallel tangent
A
O
B
{\displaystyle AOB}
and area
D
O
C
D
.
{\displaystyle DOCD.}
b
a
s
e
=
C
D
=
2.
h
e
i
g
h
t
=
H
1
T
1
=
1.
{\displaystyle \ \ \ \ \ \ base=CD=2.\ height=H_{1}T_{1}=1.}
Chord
O
C
{\displaystyle OC}
, parallel tangent
G
H
I
{\displaystyle GHI}
and area
O
H
C
O
{\displaystyle OHCO}
.
b
a
s
e
=
O
C
=
2
.
h
e
i
g
h
t
=
H
2
T
2
=
2
8
.
{\displaystyle \ \ \ \ \ \ base=OC={\sqrt {2}}.\ height=H_{2}T_{2}={\frac {\sqrt {2}}{8}}.}
See Figure 6. The curve is:
y
=
x
2
{\displaystyle y=x^{2}}
. Integral is:
x
3
3
{\displaystyle {\frac {x^{3}}{3}}}
.
Area under curve
(
O
B
C
)
{\displaystyle (OBC)}
x
=
1
=
[
x
3
3
]
=
1
3
x
=
0
{\displaystyle {\begin{aligned}x=&1\\=\ \ \ \ \ &[{\frac {x^{3}}{3}}]={\frac {1}{3}}\\x=&0\end{aligned}}}
Area under curve
(
D
O
C
)
=
{\displaystyle (DOC)=}
area
(
O
A
D
)
+
{\displaystyle (OAD)+}
area
(
O
B
C
)
=
2
3
{\displaystyle (OBC)={\frac {2}{3}}}
Area between chord
D
C
{\displaystyle DC}
and curve
D
O
C
=
2
−
2
3
=
4
3
{\displaystyle DOC=2-{\frac {2}{3}}={\frac {4}{3}}}
.
Consider the chord
C
D
{\displaystyle CD}
. Call this the
b
a
s
e
{\displaystyle base}
with value
2
{\displaystyle 2}
.
The tangent
A
O
B
{\displaystyle AOB}
through the origin is parallel to
b
a
s
e
(
D
C
)
{\displaystyle base\ (DC)}
, and the perpendicular distance
between
A
B
,
D
C
(
H
1
T
1
)
{\displaystyle AB,DC\ (H_{1}T_{1})}
is
1
{\displaystyle 1}
. Call this distance the
h
e
i
g
h
t
{\displaystyle height}
with value
1
{\displaystyle 1}
.
In this case the area enclosed between chord
D
C
{\displaystyle DC}
and curve
D
O
C
=
2
3
(
b
a
s
e
)
(
h
e
i
g
h
t
)
=
2
3
(
2
)
(
1
)
=
4
3
,
{\displaystyle DOC={\frac {2}{3}}(base)(height)={\frac {2}{3}}(2)(1)={\frac {4}{3}},}
the same as that calculated earlier.
Consider the chord
O
C
{\displaystyle OC}
and curve
O
H
C
.
{\displaystyle OHC.}
By inspection the area between
chord
O
C
{\displaystyle OC}
and curve
O
H
C
=
1
2
−
1
3
=
1
6
.
{\displaystyle OHC={\frac {1}{2}}-{\frac {1}{3}}={\frac {1}{6}}.}
Chord
O
C
{\displaystyle OC}
has equation
y
=
x
;
x
−
y
=
0
;
x
2
−
y
2
=
0
{\displaystyle y=x;\ x-y=0;\ {\frac {x}{\sqrt {2}}}-{\frac {y}{\sqrt {2}}}=0}
in normal form.
The line
G
H
I
{\displaystyle GHI}
is parallel to
b
a
s
e
O
C
{\displaystyle base\ OC}
and touches the curve at
H
(
1
2
,
1
4
)
.
{\displaystyle H({\frac {1}{2}},\ {\frac {1}{4}}).}
Distance from
H
{\displaystyle H}
to chord
O
C
=
H
2
T
2
=
1
/
2
2
−
1
/
4
2
=
1
4
2
=
2
8
=
h
e
i
g
h
t
.
{\displaystyle OC=H_{2}T_{2}={\frac {1/2}{\sqrt {2}}}-{\frac {1/4}{\sqrt {2}}}={\frac {1}{4{\sqrt {2}}}}={\frac {\sqrt {2}}{8}}=height.}
Length of
O
C
=
2
=
b
a
s
e
.
{\displaystyle OC={\sqrt {2}}=base.}
Area between chord
O
C
{\displaystyle OC}
and curve
O
H
C
=
2
3
(
b
a
s
e
)
(
h
e
i
g
h
t
)
=
2
3
(
2
)
(
2
8
)
=
2
3
(
1
4
)
=
1
6
,
{\displaystyle OHC={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\sqrt {2}})({\frac {\sqrt {2}}{8}})={\frac {2}{3}}({\frac {1}{4}})={\frac {1}{6}},}
the same
as that calculated earlier.
These observations suggest that the area enclosed between chord and curve
=
2
3
(
b
a
s
e
)
(
h
e
i
g
h
t
)
.
{\displaystyle ={\frac {2}{3}}(base)(height).}
Figure 7: The Parabola:
y
=
K
x
2
{\displaystyle y=Kx^{2}}
Origin at point
O
:
(
0
,
0
)
{\displaystyle O:(0,0)}
Points
D
,
I
:
(
p
,
0
)
,
(
p
,
K
p
2
)
{\displaystyle D,\ I:(p,0),(p,Kp^{2})}
Points
E
,
J
:
(
q
,
0
)
,
(
q
,
K
q
2
)
{\displaystyle E,\ J:(q,0),(q,Kq^{2})}
Chord
I
J
{\displaystyle IJ}
, parallel tangent
F
G
{\displaystyle FG}
and area
I
O
J
I
{\displaystyle IOJI}
, the area enclosed between chord
I
J
{\displaystyle IJ}
and curve. Length
I
J
=
b
a
s
e
.
{\displaystyle IJ=base.}
Length
H
T
=
h
e
i
g
h
t
.
{\displaystyle HT=height.}
We prove this identity for the general case. See Figure 7.
Slope of chord
I
J
=
K
q
q
−
K
p
p
q
−
p
=
K
(
(
q
+
p
)
(
q
−
p
)
)
q
−
p
=
K
(
q
+
p
)
.
{\displaystyle IJ={\frac {Kqq-Kpp}{q-p}}={\frac {K((q+p)(q-p))}{q-p}}=K(q+p).}
Find equation of chord
I
J
.
{\displaystyle IJ.}
y
=
K
(
p
+
q
)
x
+
c
;
∴
c
=
y
−
K
(
p
+
q
)
x
=
K
q
q
−
K
(
p
+
q
)
q
=
−
K
p
q
{\displaystyle y=K(p+q)x+c;\ \therefore c=y-K(p+q)x=Kqq-K(p+q)q=-Kpq}
Equation of chord
I
J
:
{\displaystyle IJ:}
y
=
K
(
p
+
q
)
x
−
K
p
q
.
{\displaystyle y=K(p+q)x-Kpq.}
Find equation of tangent
F
G
.
{\displaystyle FG.}
y
=
K
x
2
y
=
K
(
p
+
q
)
x
+
c
∴
K
x
2
=
K
(
p
+
q
)
x
+
c
K
x
2
−
K
(
p
+
q
)
x
−
c
=
0
{\displaystyle {\begin{aligned}&y=Kx^{2}\\&y=K(p+q)x+c\\\therefore \ &Kx^{2}=K(p+q)x+c\\&Kx^{2}-K(p+q)x-c=0\end{aligned}}}
We choose a value of
c
{\displaystyle c}
that gives
x
{\displaystyle x}
exactly one value.
Therefore discriminant
K
2
(
p
+
q
)
2
+
4
K
c
=
0
;
c
=
−
K
(
p
+
q
)
2
4
.
{\displaystyle K^{2}(p+q)^{2}+4Kc=0;\ c={\frac {-K(p+q)^{2}}{4}}.}
y
=
K
(
p
+
q
)
x
−
K
(
p
+
q
)
2
4
;
K
(
p
+
q
)
x
−
y
−
K
(
p
+
q
)
2
4
=
0
;
{\displaystyle y=K(p+q)x-{\frac {K(p+q)^{2}}{4}};\ K(p+q)x-y-{\frac {K(p+q)^{2}}{4}}=0;}
Equation of tangent
F
G
{\displaystyle FG}
in normal form:
K
(
p
+
q
)
x
−
y
−
K
(
p
+
q
)
2
4
K
2
(
p
+
q
)
2
+
1
=
0.
{\displaystyle {\frac {K(p+q)x-y-{\frac {K(p+q)^{2}}{4}}}{\sqrt {K^{2}(p+q)^{2}+1}}}=0.}
Equation of chord
I
J
{\displaystyle IJ}
in normal form:
K
(
p
+
q
)
x
−
y
−
K
p
q
K
2
(
p
+
q
)
2
+
1
=
0.
{\displaystyle {\frac {K(p+q)x-y-Kpq}{\sqrt {K^{2}(p+q)^{2}+1}}}=0.}
Therefore distance between chord
I
J
{\displaystyle IJ}
and tangent
F
G
{\displaystyle FG}
=
h
e
i
g
h
t
=
−
K
p
q
−
(
−
K
(
p
+
q
)
2
4
)
R
=
K
(
p
+
q
)
2
−
4
K
p
q
4
R
=
K
(
p
−
q
)
2
4
R
{\displaystyle =height={\frac {-Kpq-(-{\frac {K(p+q)^{2}}{4}})}{R}}={\frac {K(p+q)^{2}-4Kpq}{4R}}={\frac {K(p-q)^{2}}{4R}}}
where
R
=
K
2
(
p
+
q
)
2
+
1
.
{\displaystyle R={\sqrt {K^{2}(p+q)^{2}+1}}.}
Length of chord
I
J
=
b
a
s
e
=
L
=
(
K
q
q
−
K
p
p
)
2
+
(
q
−
p
)
2
.
{\displaystyle IJ=base=L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}.}
Area under chord
I
J
=
(
q
−
p
)
K
q
q
+
K
p
p
2
=
K
q
q
q
+
K
p
p
q
−
K
q
q
p
−
K
p
p
p
2
{\displaystyle IJ=(q-p){\frac {Kqq+Kpp}{2}}={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}}
Area under curve
I
O
J
{\displaystyle IOJ}
x
=
q
=
[
K
x
3
3
]
=
K
q
q
q
−
K
p
p
p
3
x
=
p
{\displaystyle {\begin{aligned}x=&q\\=\ \ \ \ \ &[{\frac {Kx^{3}}{3}}]={\frac {Kqqq-Kppp}{3}}\\x=&p\end{aligned}}}
Area between chord
I
J
{\displaystyle IJ}
and curve
I
O
J
{\displaystyle IOJ}
=
K
q
q
q
+
K
p
p
q
−
K
q
q
p
−
K
p
p
p
2
−
K
q
q
q
−
K
p
p
p
3
=
K
q
q
q
+
3
K
p
p
q
−
3
K
p
q
q
−
K
p
p
p
6
{\displaystyle ={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}-{\frac {Kqqq-Kppp}{3}}={\frac {Kqqq+3Kppq-3Kpqq-Kppp}{6}}}
=
K
(
q
−
p
)
3
6
=
K
S
6
.
{\displaystyle ={\frac {K(q-p)^{3}}{6}}={\frac {KS}{6}}.}
The aim is to prove that:
2
3
(
b
a
s
e
)
(
h
e
i
g
h
t
)
=
K
S
6
{\displaystyle {\frac {2}{3}}(base)(height)={\frac {KS}{6}}}
or
2
3
(
L
)
(
K
(
p
−
q
)
2
4
R
)
=
K
S
6
{\displaystyle {\frac {2}{3}}(L)({\frac {K(p-q)^{2}}{4R}})={\frac {KS}{6}}}
or
2
L
(
p
−
q
)
2
12
R
=
S
6
{\displaystyle {\frac {2L(p-q)^{2}}{12R}}={\frac {S}{6}}}
or
L
(
p
−
q
)
2
=
R
(
q
−
p
)
3
{\displaystyle L(p-q)^{2}=R(q-p)^{3}}
or
L
=
R
(
q
−
p
)
{\displaystyle L=R(q-p)}
where:
L
=
(
K
q
q
−
K
p
p
)
2
+
(
q
−
p
)
2
=
(
q
−
p
)
2
(
K
2
(
q
+
p
)
2
+
1
)
=
(
q
−
p
)
K
2
(
q
+
p
)
2
+
1
{\displaystyle L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}={\sqrt {(q-p)^{2}(K^{2}(q+p)^{2}+1)}}=(q-p){\sqrt {K^{2}(q+p)^{2}+1}}}
R
=
K
2
(
p
+
q
)
2
+
1
{\displaystyle R={\sqrt {K^{2}(p+q)^{2}+1}}}
R
H
S
=
(
q
−
p
)
R
=
(
q
−
p
)
K
2
(
p
+
q
)
2
+
1
=
L
.
{\displaystyle RHS=(q-p)R=(q-p){\sqrt {K^{2}(p+q)^{2}+1}}=L.}
Therefore
L
=
R
(
q
−
p
)
{\displaystyle L=R(q-p)}
and area enclosed between curve and chord =
2
3
(
b
a
s
e
)
(
h
e
i
g
h
t
)
{\displaystyle {\frac {2}{3}}(base)(height)}
where
b
a
s
e
{\displaystyle base}
is the length of the chord, and
h
e
i
g
h
t
{\displaystyle height}
is the perpendicular distance between chord and tangent parallel to chord.
Consider parabola:
16
x
2
−
24
x
y
+
9
y
2
+
20
x
−
140
y
+
600
=
0
{\displaystyle 16x^{2}-24xy+9y^{2}+20x-140y+600=0}
and chord:
4
x
+
3
y
−
96
=
0.
{\displaystyle 4x+3y-96=0.}
The aim is to calculate area between chord and curve.
Calculate the points at which chord and parabola intersect:
(
5
7
16
,
24
3
4
)
,
(
15
1
3
,
11
5
9
)
.
{\displaystyle (5{\frac {7}{16}},24{\frac {3}{4}}),\ (15{\frac {1}{3}},11{\frac {5}{9}}).}
Method 1. By chord and parallel tangent
edit
Figure 8: The Parabola:
16
x
2
−
24
x
y
+
9
y
2
+
20
x
−
140
y
+
600
=
0
{\displaystyle 16x^{2}-24xy+9y^{2}+20x-140y+600=0}
Chord
A
D
:
4
x
+
3
y
−
96
=
0.
{\displaystyle AD:4x+3y-96=0.}
Length
A
D
=
b
a
s
e
=
2375
144
.
{\displaystyle AD=base={\frac {2375}{144}}.}
Parallel tangent
C
F
B
.
h
e
i
g
h
t
=
B
E
=
361
24
.
{\displaystyle CFB.\ height=BE={\frac {361}{24}}.}
Area between chord and curve
=
D
E
A
F
D
.
{\displaystyle =DEAFD.}
Length of chord
=
(
15
1
3
−
5
7
16
)
2
+
(
11
5
9
−
24
3
4
)
2
=
(
46
3
−
87
16
)
2
+
(
99
4
−
104
9
)
2
=
(
475
48
)
2
+
(
475
36
)
2
=
(
475
(
3
)
48
(
3
)
)
2
+
(
475
(
4
)
36
(
4
)
)
2
=
(
475
(
3
)
144
)
2
+
(
475
(
4
)
144
)
2
=
475
(
5
)
144
=
2375
144
=
b
a
s
e
.
{\displaystyle {\begin{aligned}=&\ {\sqrt {(15{\frac {1}{3}}-5{\frac {7}{16}})^{2}+(11{\frac {5}{9}}-24{\frac {3}{4}})^{2}}}\\\\=&\ {\sqrt {({\frac {46}{3}}-{\frac {87}{16}})^{2}+({\frac {99}{4}}-{\frac {104}{9}})^{2}}}\\\\=&\ {\sqrt {({\frac {475}{48}})^{2}+({\frac {475}{36}})^{2}}}=\ {\sqrt {({\frac {475(3)}{48(3)}})^{2}+({\frac {475(4)}{36(4)}})^{2}}}\\\\=&\ {\sqrt {({\frac {475(3)}{144}})^{2}+({\frac {475(4)}{144}})^{2}}}=\ {\frac {475(5)}{144}}\\\\=&\ {\frac {2375}{144}}=base.\end{aligned}}}
Equation of chord in normal form:
4
5
x
+
3
5
y
−
19.2
=
0.
{\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y-19.2=0.}
Equation of parallel tangent in normal form:
4
5
x
+
3
5
y
+
g
=
0
,
{\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y+g=0,}
where
g
=
4
A
F
e
e
−
4
B
F
d
e
+
4
C
F
d
d
−
D
D
e
e
+
2
D
E
d
e
−
E
E
d
d
4
A
E
e
−
2
B
D
e
−
2
B
E
d
+
4
C
D
d
{\displaystyle g={\frac {4AFee-4BFde+4CFdd-DDee+2DEde-EEdd}{4AEe-2BDe-2BEd+4CDd}}}
and
d
=
4
5
,
e
=
3
5
{\displaystyle d={\frac {4}{5}},\ e={\frac {3}{5}}}
and
A
=
16
,
B
=
−
24
,
C
=
9
,
D
=
20
,
E
=
−
140
,
F
=
600.
{\displaystyle A=16,\ B=-24,\ C=9,\ D=20,\ E=-140,\ F=600.}
g
=
−
499
120
.
{\displaystyle g=-{\frac {499}{120}}.}
Equation of chord in normal form:
4
5
x
+
3
5
y
−
19.2
=
0.
{\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y-19.2=0.}
Equation of parallel tangent in normal form:
4
5
x
+
3
5
y
−
499
120
=
0.
{\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y-{\frac {499}{120}}=0.}
Distance between chord and parallel tangent
=
−
499
120
−
(
−
19.2
)
=
361
24
=
h
e
i
g
h
t
.
{\displaystyle =-{\frac {499}{120}}-(-19.2)={\frac {361}{24}}=height.}
Area enclosed between chord and curve
=
2
3
(
b
a
s
e
)
(
h
e
i
g
h
t
)
=
2
3
(
2375
144
)
(
361
24
)
=
(
2375
)
(
361
)
3
(
12
3
)
.
{\displaystyle ={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\frac {2375}{144}})({\frac {361}{24}})={\frac {(2375)(361)}{3(12^{3})}}.}
Method 2. By identifying the basic parabola.
edit