The quadratic function has maximum power of $x$ equal to $2$ :

$Ax^{2}+Bx+C$ .

$Ax^{2}+Bx+C=0\ \dots \ (1)$ , in which coefficient $A$ is non-zero.

The solution of the quadratic equation is:

$x={\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\ \dots \ (2)$ The above solution to the quadratic is well known to high-school algebra students. The proof of the solution is usually presented to students as completion of the square, not presented here. To me completion of the square seemed too complicated. To help you acquire a new respect for the quadratic, presented below are more ways to solve it.

The depressed cubic leads to a solution of the cubic and the depressed quartic leads to a solution of the quartic. This paragraph shows that the depressed quadratic leads to a solution of the quadratic.

To produce the depressed function, let

$x={\frac {-B+t}{GA}}$

where G is the degree of the function. For the quadratic G = 2. Let

$x={\frac {-B+t}{2A}}\ \dots \ (3)$

Substitute (3) into (1) and expand:

$A({\frac {-B+t}{2A}})({\frac {-B+t}{2A}})+B({\frac {-B+t}{2A}})+C\ \dots \ (4)$

$(4)\ *\ 4A,\ 4AA({\frac {-B+t}{2A}})({\frac {-B+t}{2A}})+4AB({\frac {-B+t}{2A}})+4AC$

$(-B+t)(-B+t)+2B(-B+t)+4AC$

$B^{2}-2Bt+t^{2}-2B^{2}+2Bt+4AC$

$t^{2}-B^{2}+4AC$

In the depressed quadratic above the coefficient of $t^{2}$  is $1$  and that of $t$  is $0$ .

$t^{2}=B^{2}-4AC$

$t=\pm {\sqrt {B^{2}-4AC}}\ \dots \ (5)$

Substitute (5) into (3) and the result is the solution in (2).

p +- q

Let one value of $x$  be $p+q$  and another value of $x$  be $p-q$ . Substitute these values into $(1)$  above and expand.

{\begin{aligned}A(p+q)(p+q)+B(p+q)+C=&\ 0\\App+A2pq+Aqq+Bp+Bq+C=&\ 0\ \dots \ (6)\\\\A(p-q)(p-q)+B(p-q)+C=&\ 0\\App-A2pq+Aqq+Bp-Bq+C=&\ 0\ \dots \ (7)\end{aligned}}

{\begin{aligned}(6)-(7),\ 4Apq+2Bq=&\ 0\\2Ap+B=&\ 0\\2Ap=&\ -B\\\\p=&\ {\frac {-B}{2A}}\ \dots \ (8)\end{aligned}}

{\begin{aligned}(6)+(7),\ 2App+2Aqq+2Bp+2C=&\ 0\\App+Aqq+Bp+C=&\ 0\\App+Bp+C=&\ -Aqq\ \dots \ (9)\\\\(9)*4A,\ 4AApp+4ABp+4AC=&\ -4AAqq\ \dots \ (10)\end{aligned}}

Substitute $(8)$  into $(10)$

{\begin{aligned}(-B)(-B)+2B(-B)+4AC=&\ -4AAqq\\BB-2BB+4AC=&\ -4AAqq\\4AAqq=&\ BB-4AC\\\\qq=&\ {\frac {BB-4AC}{4AA}}\\\\q=&\ {\frac {\pm {\sqrt {B^{2}-4AC}}}{2A}}\ \dots \ (11)\end{aligned}}

{\begin{aligned}x=&\ p+q\\\\=&\ ({\frac {-B}{2A}})+({\frac {\pm {\sqrt {B^{2}-4AC}}}{2A}})\\\\=&\ {\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\end{aligned}}

By observation and elementary deduction

You have drawn a few curves by hand and suppose that each curve is symmetrical about the vertical line through a minimum point. Given $f(x)=Ax^{2}+Bx+C$  you suppose that $f(p+q)=f(p-q)$ .

{\begin{aligned}A(p+q)(p+q)+B(p+q)+C=&\ A(p-q)(p-q)+B(p-q)+C\\\\A(pp+2pq+qq)+Bp+Bq=&\ A(pp-2pq+qq)+Bp-Bq\\\\A2pq+Bq=&\ -A2pq-Bq\\\\2A2pq+2Bq=&\ 0\\\\2q(2Ap+B)=&\ 0\\\\2Ap+B=&\ 0\\\\p=&\ {\frac {-B}{2A}}\end{aligned}}

You have found the stationary point without using calculus. Continue as per calculus below.

By calculus

The derivative of $(1)$  is $2Ax+B$  which equals $0$  at a stationary point.

At the stationary point $x={\frac {-B}{2A}}$ .

Prove that the function is symmetrical about the vertical line through $x={\frac {-B}{2A}}$ .

Let $x={\frac {-B}{2A}}+p\ \dots \ (12)$

Substitute (12) in (1) and expand:

${\frac {4AApp+4AC-BB}{4A}}$

Let $x={\frac {-B}{2A}}-p\ \dots \ (13)$

Substitute (13) in (1) and expand

${\frac {4AApp+4AC-BB}{4A}}$

The expansion is the same for both values of x, (12) and (13). The curve is symmetrical.

If the function equals 0, then

{\begin{aligned}+4AApp+4AC-BB=&\ 0\\\\4AApp=&\ BB-4AC\\\\pp=&\ {\frac {BB-4AC}{4AA}}\\\\p=&\ {\frac {\pm {\sqrt {B^{2}-4AC}}}{2A}}\\\end{aligned}}

Substitute this value of p in (12) and the result is (2).

By movement of the vertex

Begin with the basic quadratic: $y=Ax^{2}$ .

Move vertex from origin $(0,0)$  to $(h,k)$ .

$y-k=A(x-h)^{2}$ .

$y-k=A(x^{2}-2hx+h^{2})$ .

$y-k=Ax^{2}-2Ahx+Ah^{2}$ .

$y=Ax^{2}-2Ahx+Ah^{2}+k$ .

This equation is in the form of the quadratic $y=Ax^{2}+Bx+C$  where:

$B=-2Ah;\ C=Ah^{2}+k$ .

Therefore $h={\frac {-B}{2A}}$  and $h$  is the X coordinate of the vertex in the new position.

Continue as per calculus above.

p + qi

Let $x=p+qi$  where $i={\sqrt {-1}}$

Substitute this value of $x$  into (1) and expand:

{\begin{aligned}A(p+qi)(p+qi)+B(p+qi)+C\\\\A(p^{2}+2pqi+(qi)^{2})+Bp+Bqi+C\\\\Ap^{2}+2Apqi-Aq^{2}+Bp+Bqi+C\end{aligned}}

Terms containing $i=2Apqi+Bqi\ \dots \ (14)$

From (14), $2Ap+B=0$  and $p={\frac {-B}{2A}}\ \dots \ (15)$

Terms without $i=Ap^{2}-Aq^{2}+Bp+C\ \dots \ (16)$

From (15) and (16):

{\begin{aligned}Aq^{2}=&\ Ap^{2}+Bp+C\\\\=&\ A({\frac {-B}{2A}})({\frac {-B}{2A}})+B({\frac {-B}{2A}})+C\\\\q^{2}=&\ {\frac {B^{2}}{4AA}}-{\frac {B^{2}}{2AA}}+{\frac {C}{A}}\\\\=&\ {\frac {B^{2}}{4AA}}-{\frac {2B^{2}}{4AA}}+{\frac {4AC}{4AA}}\\\\=&\ {\frac {-B^{2}+4AC}{4AA}}\\\\=&\ {\frac {-1(B^{2}-4AC)}{4AA}}\\\\q=&\ {\frac {\pm \ i{\sqrt {B^{2}-4AC}}}{2A}}\\\\x=&\ (p)+(q)i\\\\=&\ ({\frac {-B}{2A}})\pm ({\frac {i{\sqrt {B^{2}-4AC}}}{2A}})i\\\\=&\ {\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\end{aligned}}

This method shows the imaginary value $i$  coming into existence to help with intermediate calculations and then going away before the end result appears.

The quadratic function is usually defined as the familiar polynomial (1). However, the quadratic may be defined in other ways. Some are shown below:

By three points

If three points on the curve are known, the familiar polynomial may be deduced. For example, if three points $(-5,40),\ (4,13),\ (7,76)$  are given and the three points satisfy $y=f(x)$ , the values $A,B,C$  may be calculated.

{\begin{aligned}A(-5)(-5)+B(-5)+C=&\ 40\\A(4)(4)+B(4)+C=&\ 13\\A(7)(7)+B(7)+C=&\ 76\\25A-5B+C=&\ 40\ \dots \ (17)\\16A+4B+C=&\ 13\ \dots \ (18)\\49A+7B+C=&\ 76\ \dots \ (19)\\\end{aligned}}

The solution of the three equations $(17),\ (18),\ (19)$  gives the equation $y=2x^{2}-x-15$ .

If the three points were to satisfy $x=f(y)$ , the equation would be $x={\frac {2}{189}}y^{2}-{\frac {169}{189}}y+{\frac {2615}{189}}$ .

By two points and a slope

Given two points $(-4,13)$  and $(1,48)$  and the slope at $(1,48)=32$ , calculate $A,B,C$ .

{\begin{aligned}A(-4)(-4)+B(-4)+C=\ 13\\A(1)(1)+B(1)+C=\ 48\\\end{aligned}}

Slope = 2Ax + B, therefore

{\begin{aligned}2A(1)+B=&\ 32\\\\16A-4B+C=&\ 13\ \dots \ (20)\\A+B+C=&\ 48\ \dots \ (21)\\2A+B=&\ 32\ \dots \ (22)\\\end{aligned}}

The solution of the three equations $(20),(21),(22)$  gives the polynomial $5x^{2}+22x+21\ \dots \ (23)$ .

By movement of the vertex

Begin with the basic quadratic $y=x^{2}$ .

If $x$  has the value $p$ , then $y=p^{2}$  and the height of $y$  above the vertex = $p^{2}$ .

If we move the vertex to $(h,k)$ , then the equation becomes $(y-k)=(x-h)^{2}$ .

If $x$  has the value $h+p$ , then $y-k=(h+p-h)^{2}=p^{2};\ y=p^{2}+k$

and the height of $y$  above the vertex = $y-k=p^{2}+k-k=p^{2}$ .

The curve $y=x^{2}$  and the curve $y-k=(x-h)^{2}$  have the same shape.

It's just that the vertex of the former $(0,0)$  has been moved to the vertex of the latter $(h,k)$ .

The latter equation expanded becomes $y=x^{2}-2hx+h^{2}+k=x^{2}+(-2h)x+(h^{2}+k)$ .

The example under "two points and a slope" above is $y=5x^{2}+22x+21\ \dots \ (23)$ .

Therefore

{\begin{aligned}y=&\ ax^{2}\\\\y-k=&\ a(x-h)^{2}\\\\y-k=&\ a(x^{2}-2hx+h^{2})=ax^{2}-2ahx+ah^{2}\\\\y=&\ ax^{2}-2ahx+ah^{2}+k\\\\a=&\ 5\\\\-2ah=&\ 22;\ h={\frac {22}{-2(5)}}=-2.2\\\\ah^{2}+k=&\ 21\\\\k=&\ 21-5(-2.2)(-2.2)=-3.2\\\\\end{aligned}}

The example $(23)$  may be expressed as

{\begin{aligned}y-(-3.2)=&\ 5(x-(-2.2))^{2}\\\\y+3.2=&\ 5(x+2.2)^{2}\end{aligned}}

For proof, expand:

{\begin{aligned}y+3.2=&\ 5(x+2.2)^{2}\\=&\ 5(x^{2}+4.4x+4.84)\\=&\ 5x^{2}+22x+24.2\\y=&\ 5x^{2}+22x+24.2-3.2=5x^{2}+22x+21\end{aligned}}

By compliance with the standard equation of the conic section

The quadratic function can comply with the format: $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.$  (See The General Quadratic below.)

For example, the function $y=3x^{2}+5x-7$  can be expressed as:

$(3)x^{2}+(0)xy+(0)y^{2}+(5)x+(-1)y+(-7)=0$  or:

$3x^{2}+5x-7-y=0.$

To express a valid quadratic in this way, both $A,E$  or both $C,D$  must be non-zero.

By a point and a straight line

The point is called the focus and the line is called the directrix. The distance from point to line is non-zero. The quadratic is the locus of a point that is equidistant from both focus and line at all times. When the quadratic is defined in this way, it is usually called a parabola.

Let the $focus$  have coordinates $(p,q).$

Let the $directrix$  have equation: $y=k.$

Let the point $(x,y)$  be equidistant from both focus and directrix.

Distance from $(x,y)$  to focus $={\sqrt {(x-p)^{2}+(y-q)^{2}}}$ .

Distance from $(x,y)$  to directrix $=y-k$ .

By definition these two lengths are equal.

${\sqrt {(x-p)^{2}+(y-q)^{2}}}=y-k$

$x^{2}-2px+p^{2}+y^{2}-2qy+q^{2}=y^{2}-2ky+k^{2}$

$(2q-2k)y=x^{2}-2px+p^{2}+q^{2}-k^{2}$

$y={\frac {x^{2}}{2q-2k}}+{\frac {-2px}{2q-2k}}+{\frac {p^{2}+q^{2}-k^{2}}{2q-2k}}$

Let this equation have the form: $y=Ax^{2}+Bx+C$

Therefore:

{\begin{aligned}A&={\frac {1}{2q-2k}}\\B&={\frac {-2p}{2q-2k}}\\C&={\frac {p^{2}+q^{2}-k^{2}}{2q-2k}}\end{aligned}}

Given $A,B,C$  calculate $p,q,k$ .

$B={\frac {-2p}{2q-2k}}=-2pA;\ p={\frac {-B}{2A}}$

There are two equations with two unknowns $q,k:$

{\begin{aligned}A&(2q-2k)-1=0\\C&(2q-2k)-p^{2}-q^{2}+k^{2}=0\end{aligned}}

The solutions are:

{\begin{aligned}p&={\frac {-B}{2A}}\\q&={\frac {1-(B^{2}-4AC)}{4A}}\\k&={\frac {-1-(B^{2}-4AC)}{4A}}\end{aligned}}

$y=x^{2}-2x-3$  showing :
* X and Y intercepts in red,
* vertex and axis of symmetry in blue,
* focus and directrix in pink.

If the quadratic equation is expressed as $y=Ax^{2}+Bx+C$  then:

The focus is the point $({\frac {-B}{2A}},{\frac {1-(B^{2}-4AC)}{4A}})$ , and

The directrix has equation: $y={\frac {-1-(B^{2}-4AC)}{4A}}$ .

The $vertex$  is exactly half-way between focus and directrix.

Vertex is the point $({\frac {-B}{2A}},{\frac {-(B^{2}-4AC)}{4A}})$ .

$A={\frac {1}{2q-2k}};\ 2q-2k={\frac {1}{A}};\ q-k={\frac {1}{2A}}=$  distance from directrix to focus.

Distance from vertex to focus $={\frac {1}{4A}}$ .

If the curve has equation $y=Ax^{2}$ , then the vertex is at the origin $(0,0)$ .

If the focus is the point $(0,q)$ , then $q={\frac {1}{4A}};\ A={\frac {1}{4q}}$  and the equation $y=Ax^{2}$  becomes $y={\frac {x^{2}}{4q}}$ .

\displaystyle {\begin{aligned}\end{aligned}}
\displaystyle {\begin{aligned}\end{aligned}}

An example with vertical focus

Figure 3: Graph of quadratic function with vertical focus $8(y+1)=(x-4)^{2}$  showing :
* vertex at (4,-1),
* focus at (4,1),
* directrix at y = -3.

Let $(p,q)=(4,1),\ k=-3.$

Directrix has equation: $y=-3$ . Focus has coordinates $(4,1)$ .

{\begin{aligned}A&={\frac {1}{2q-2k}}={\frac {1}{2(1)-2(-3)}}={\frac {1}{2+6}}={\frac {1}{8}}\\B&={\frac {-2p}{2q-2k}}={\frac {-2(4)}{8}}=-1\\C&={\frac {p^{2}+q^{2}-k^{2}}{2q-2k}}={\frac {16+1-9}{8}}=1\end{aligned}}

This example has equation: $y={\frac {1}{8}}x^{2}-x+1$  or $8y=x^{2}-8x+8$  or $8(y+1)=(x-4)^{2}$ . See Figure 3.

Distance from vertex to focus = ${\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}={\frac {8}{4}}=2.$

Or:

Vertex has coordinates $(4,-1).$

Distance from vertex to focus $=2={\frac {1}{4A}};\ A={\frac {1}{8}}$ .

Curve has shape of $y={\frac {1}{8}}x^{2}$  with vertex moved to $(4,-1).\ y-(-1)={\frac {1}{8}}(x-4)^{2};\ 8(y+1)=(x-4)^{2}.$

Let the $focus$  have coordinates $(p,q).$

Let the $directrix$  have equation: $x=k.$

Let the point $(x,y)$  be equidistant from both focus and directrix.

Distance from $(x,y)$  to focus $={\sqrt {(x-p)^{2}+(y-q)^{2}}}$ .

Distance from $(x,y)$  to directrix $=x-k$ .

By definition these two lengths are equal.

${\sqrt {(x-p)^{2}+(y-q)^{2}}}=x-k$

$x^{2}-2px+p^{2}+y^{2}-2qy+q^{2}=x^{2}-2kx+k^{2}$

$(2p-2k)x=y^{2}-2qy+p^{2}+q^{2}-k^{2}$

$x={\frac {y^{2}}{2p-2k}}+{\frac {-2qy}{2p-2k}}+{\frac {p^{2}+q^{2}-k^{2}}{2p-2k}}$

Let this equation have the form: $x=Ay^{2}+By+C$

Therefore:

{\begin{aligned}A&={\frac {1}{2p-2k}}\\B&={\frac {-2q}{2p-2k}}\\C&={\frac {p^{2}+q^{2}-k^{2}}{2p-2k}}\end{aligned}}

Given $A,B,C$  calculate $p,q,k.$

$B={\frac {-2q}{2p-2k}}=-2qA;\ q={\frac {-B}{2A}}$

There are two equations with two unknowns $p,k:$

{\begin{aligned}A&(2p-2k)-1=0\\C&(2p-2k)-p^{2}-q^{2}+k^{2}=0\end{aligned}}

The solutions are:

{\begin{aligned}p&={\frac {1-(B^{2}-4AC)}{4A}}\\q&={\frac {-B}{2A}}\\k&={\frac {-1-(B^{2}-4AC)}{4A}}\end{aligned}}

If the quadratic equation is expressed as $x=Ay^{2}+By+C$  then:

The focus is the point $({\frac {1-(B^{2}-4AC)}{4A}},{\frac {-B}{2A}})$ , and

The directrix has equation: $x={\frac {-1-(B^{2}-4AC)}{4A}}$ .

The $vertex$  is exactly half-way between focus and directrix.

Vertex is the point $({\frac {-(B^{2}-4AC)}{4A}},{\frac {-B}{2A}})$ .

$A={\frac {1}{2p-2k}};\ 2p-2k={\frac {1}{A}};\ p-k={\frac {1}{2A}}=$  distance from directrix to focus.

Distance from vertex to focus $={\frac {1}{4A}}$ .

If the curve has equation $x=Ay^{2}$ , then the vertex is at the origin $(0,0)$ .

If the focus is the point $(p,0)$ , then $p={\frac {1}{4A}};\ A={\frac {1}{4p}}$  and the equation $x=Ay^{2}$  becomes $x={\frac {y^{2}}{4p}}$ .

An example with horizontal focus

Figure 4: Graph of quadratic function with horizontal focus $8(x+1)=(y-4)^{2}$  showing :
* vertex at (-1,4),
* focus at (1,4),
* directrix at x = -3.

Let $(p,q)=(1,4),\ k=-3.$

Directrix has equation: $x=-3$ . Focus has coordinates $(1,4)$ .

{\begin{aligned}A&={\frac {1}{2p-2k}}={\frac {1}{2(1)-2(-3)}}={\frac {1}{8}}\\B&={\frac {-2q}{2p-2k}}={\frac {-2(4)}{8}}=-1\\C&={\frac {p^{2}+q^{2}-k^{2}}{2p-2k}}={\frac {1+16-9}{8}}=1\end{aligned}}

This example has equation: $x={\frac {1}{8}}y^{2}-y+1$  or $8x=y^{2}-8y+8$  or $8(x+1)=(y-4)^{2}$ . See Figure 4.

Distance from vertex to focus = ${\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}={\frac {8}{4}}=2.$

Given equation $8x=y^{2}-8y+8$  calculate $p,q,k$ .

Method 1. By algebra

Put equation in form: $x={\frac {1}{8}}y^{2}-y+1$  where $A={\frac {1}{8}},\ B=-1,\ C=1.$

{\begin{aligned}p&={\frac {1-(B^{2}-4AC)}{4A}}={\frac {1-(1-{\frac {1}{2}})}{4({\frac {1}{8}})}}={\frac {1-{\frac {1}{2}}}{\frac {1}{2}}}=1\\q&={\frac {-B}{2A}}={\frac {-(-1)}{2({\frac {1}{8}})}}={\frac {8}{2}}=4\\k&={\frac {-1-(B^{2}-4AC)}{4A}}={\frac {-1-(1-{\frac {1}{2}})}{\frac {1}{2}}}={\frac {-1{\frac {1}{2}}}{\frac {1}{2}}}=-3\end{aligned}}

Method 2. By analytical geometry

Distance from vertex to focus $={\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}=2.$

Put equation in $semi$ -$reduced$  form:

{\begin{aligned}8x=y^{2}-8y+8\\(y-4)^{2}=y^{2}-8y+16\\8x=(y-4)^{2}-8\\8x+8=(y-4)^{2}\\8(x+1)=(y-4)^{2}\end{aligned}}

Vertex is point $(-1,4).$

Focus is point $(-1+2,4)=(1,4)=(p,q).$

Directrix has equation: $x=-1-2=-3=k.$

## The Parabola

Figure 1: The Parabola $y={\frac {x^{2}}{4p}}$
Focus at point $F\ (0,p)$
Vertex at origin $(0,0)$
Directrix is line $y=-p$
By definition $BF=BH$  and $DF=DG$
In Figure 1 $p=1$

See Figure 1. For simplicity values are defined as shown. Let an arbitrary point on the curve be $(x,y)$ .

By definition, ${\sqrt {(x-0)^{2}+(y-p)^{2}}}=y+p$ . This expression expanded gives:

$x^{2}-4py=0;\ y={\frac {x^{2}}{4p}}$  and slope = ${\frac {x}{2p}}$ .

If the equation of the curve is expressed as: $y=Kx^{2}$ , then $K={\frac {1}{4p}};\ 4Kp=1;\ p={\frac {1}{4K}}$ .

Let a straight line through the focus intersect the parabola in two points $(x_{1},y_{1})$  and $(x_{2},y_{2})$ .

{\begin{aligned}y=&\ {\frac {x^{2}}{4p}}=mx+p\\\\x^{2}=&\ 4pmx+4pp\\\\x^{2}-4pmx-4pp=&\ 0\\\\x=&\ {\frac {4pm\pm {\sqrt {16ppmm+16pp}}}{2}}\\\\=&\ {\frac {4pm\pm 4p{\sqrt {mm+1}}}{2}}\\\\=&\ 2pm\pm 2p{\sqrt {mm+1}}\\\\=&\ 2pm\pm 2pR\\\end{aligned}}

where

$m$  is the slope of line DB in Figure 1.

{\begin{aligned}R=&\ {\sqrt {m^{2}+1}}\\\\x_{1}=&\ 2pm+2pR\\\\x_{2}=&\ 2pm-2pR\\\\y_{1}=&\ {\frac {x_{1}^{2}}{4p}}=2Rmp+2mmp+p\\\\y_{2}=&\ -2Rmp+2mmp+p\end{aligned}}

Characteristics of the Parabola

The parabola is a grab-bag of many interesting facts. We prove first that the tangents at $(x_{1},y_{1})$  and $(x_{2},y_{2})$  are perpendicular.

{\begin{aligned}s=&\ {\frac {x}{2p}}\\\\s1=&\ {\frac {x_{1}}{2p}}=m+R\\\\s2=&\ m-R\end{aligned}}

The product of $s_{1}$  and $s_{2}=(m+R)(m-R)=m^{2}-R^{2}=m^{2}-(m^{2}+1)=-1$ . Therefore, the tangents (lines AB and AD in Figure 1) are perpendicular.

Second, we prove that the two tangents intersect on the directrix. Using $y=mx+c$  and $c=y-mx$ :

{\begin{aligned}c_{1}=&\ y_{1}-(s_{1})(x_{1})\\\\=&\ 2Rmp+2mmp+p-(m+r)(2pm+2pR)\\\\=&\ -2Rmp-2mmp-p\\\\c_{2}=&\ 2Rmp-2mmp-p\end{aligned}}

The $y$  coordinate of the point of intersection satisfies both $s_{1}x+c_{1}$  and $s_{2}x+c_{2}$ . Therefore,

{\begin{aligned}(m+R)x+(-2Rmp-2mmp-p)=&\ (m-R)x+(2Rmp-2mmp-p)\\\\x=&\ 2mp\\\end{aligned}}

$2mp$  is the mid-point between $x_{1}$  and $x_{2}$ . Point A in Figure 1 has coordinates $(2mp,-p)$

Check our work:

{\begin{aligned}y=s_{1}x+c_{1}=(m+R)2mp+(-2Rmp-2mmp-p)=-p\\y=s_{2}x+c_{2}=(m-R)2mp+(+2Rmp-2mmp-p)=-p\\\end{aligned}}

The tangents intersect at $(2mp,-p)$ . They intersect on the directrix where $y=-p$ . See Tangents perpendicular and oscillating.

Third, we prove that the triangle defined by the three points $H\ (x1,-p),G\ (x2,-p)$  and $F\ (0,p)$  is a right triangle.

Slope of line $(0,p)...(x_{1},-p)=s_{3}={\frac {p-(-p)}{0-x_{1}}}={\frac {2p}{-(2pm+2pR)}}={\frac {-1}{m+R}}$ .

Slope of line $(0,p)...(x_{2},-p)=s_{4}={\frac {p-(-p)}{0-x_{2}}}={\frac {2p}{-(2pm-2pR)}}={\frac {-1}{m-R}}$ .

$(s_{3})*(s_{4})={\frac {-1}{m+R}}\ *\ {\frac {-1}{m-R}}={\frac {1}{m^{2}-R^{2}}}={\frac {1}{-1}}=-1$

The product of $s_{3}$  and $s_{4}$  is $-1$ . Therefore the two sides $HF,GF$  are perpendicular and the triangle $HFG$  in Figure 1 is a right triangle with the right angle at $F$ .

Fourth, we prove that the two lines $AF,DFB$  are perpendicular.

Point $A:\ (2mp,-p)$ . Point $F:(0,p)$ .

Using slope $={\frac {y_{1}-y_{2}}{x_{1}-x_{2}}}$

Slope of line $AF=s_{5}={\frac {p-(-p)}{0-2mp}}={\frac {2p}{-2mp}}={\frac {-1}{m}}$ .

Slope of line $DFB=s_{6}=m.$

$(s_{5})(s_{6})=-1.$  Therefore the two lines $AF,DFB$  are perpendicular.

Figure 1: The Parabola $y={\frac {x^{2}}{4p}}$
Focus at point $F\ (0,p)$
Vertex at origin $(0,0)$
Directrix is line $y=-p$
By definition $BF=BH$  and $DF=DG$
In Figure 1 $p=1$

In the last section we proved several points about the parabola, beginning with line $DFB$  and moving towards point $A$  on the directrix. In this section, we prove the reverse, beginning with point $A$  and moving towards line $DFB$ .

Let $(k,-p)$  be any point on the directrix $y=-p$ .

Using $y=mx+c,-p=mk+c,\ c=-p-mk$  and any line through $(k,-p)$  is defined as $y=sx-p-sk$  where $s$  is the slope of the line.

Let this line intersect the parabola $y={\frac {x^{2}}{4p}}$ . (In Figure 1, p = 1.)

{\begin{aligned}y&\ ={\frac {x^{2}}{4p}}=sx-p-sk\\\\x^{2}&\ =4psx-4pp-4psk\\\\x^{2}&\ -4psx+4pp+4psk=0\ \dots \ (24)\\\\x^{2}&\ +(-4ps)x+(4pp+4psk)=0\end{aligned}}

The above defines the $X$  coordinate/s of any line through $(k,-p)$  that intersect/s the parabola. We are interested in the tangent, a line that intersects in only one place. Therefore, the discriminant is 0.

{\begin{aligned}16ppss-4(4pp+4psk)=0\\\\16ppss-16(pp+psk)=0\\\\ppss-pks-pp=0\\\\pss-ks-p=0\\\\s=\ {\frac {k\pm {\sqrt {kk+4pp}}}{2p}}=\ {\frac {k\pm R}{2p}}\end{aligned}}

where $R={\sqrt {kk+4pp}}$

Slope of tangent1 = $s_{1}={\frac {k+R}{2p}}$  (In Figure 1, tangent1 is the line $ABC$ .)

Slope of tangent2 = $s_{2}={\frac {k-R}{2p}}$  (In Figure 1, tangent2 is the line $ADE$ .)

Prove that tangent1 and tangent2 are perpendicular.

$s_{1}*s_{2}={\frac {k+R}{2p}}*{\frac {k-R}{2p}}={\frac {kk-RR}{4pp}}={\frac {kk-(kk+4pp)}{4pp}}={\frac {-4pp}{4pp}}=-1$

The product of the two slopes is -1. Therefore, the two tangents are perpendicular.

From (24), we chose a value of $s$  that made the discriminant 0. Therefore

{\begin{aligned}x=&\ {\frac {-B}{2A}}={\frac {4ps}{2}}=2ps\\\\x_{1}=&\ 2ps_{1}=2p*{\frac {k+R}{2p}}=k+R\\\\x_{2}=&\ k-R\\\\y=&\ {\frac {x^{2}}{4p}}\\\\y_{1}=&\ {\frac {(k+R)^{2}}{4p}}={\frac {kk+2kR+RR}{4p}}={\frac {kk+2kR+kk+4pp}{4p}}={\frac {2kk+2kR+4pp}{4p}}={\frac {kk+kR+2pp}{2p}}\\\\y_{2}=&\ {\frac {kk-kR+2pp}{2p}}\\\\m=&\ {\frac {y_{1}-y_{2}}{x_{1}-x_{2}}}={\frac {kR/p}{2R}}={\frac {k}{2p}}\end{aligned}}

(In Figure 1, $m$  is the slope of line $DFB$ . This statement agrees with $k=2mp$  proved in the last section.)

We have a line joining the two points $(x1,y1),(x_{2},y_{2})$ . Calculate the intercept on the $Y$  axis.

Using $y=mx+c$ ,

{\begin{aligned}c=&\ y-mx=y_{1}-mx_{1}\\\\=&\ {\frac {kk+kR+2pp}{2p}}-{\frac {k}{2p}}*(k+R)\\\\=&\ {\frac {kk+kR+2pp-(kk+kR)}{2p}}\\\\=&\ {\frac {2pp}{2p}}=p\end{aligned}}

The line joining the two points $(x_{1},y_{1}),\ (x_{2},y_{2})$  passes through the focus $(0,p)$ .

Figure 2: The Parabola $y={\frac {x^{2}}{4p}}$
Lines $JGK,DFB$  are parallel.
Line $AGH$  divides area $DGBHFD$  into two halves equal by area.

Two lines parallel

In Figure 2 tangents $ABC$  and $ADE$  intersect at point $A$  on the directrix.

The line $AGH$  has value $x=k$ . The line $JGK$  is tangent to the curve at $G$ .

The slope of tangent $JGK={\frac {x}{2p}}={\frac {k}{2p}}$ . The slope of line $DFB$  also $={\frac {k}{2p}}$ .

Therefore two lines $JGK,DFB$  are parallel.

Area DGBHFD

$x_{1}=2pm+2pR$

$x_{2}=2pm-2pR$

where $R={\sqrt {m^{2}+1}}$

Line $DFB=mx+p$ . The integral of this value $=m{\frac {x^{2}}{2}}+px$ .

Area under line $DFB$

{\begin{aligned}x_{1}&\\=\ \ \ \ &[m{\frac {x^{2}}{2}}+px]=8Rmmpp+4Rpp\\x_{2}&\end{aligned}}

Area under curve $DGB$

{\begin{aligned}x_{1}&\\=\ \ \ \ &[{\frac {x^{3}}{12p}}]={\frac {16Rmmpp+4Rpp}{3}}\\x_{2}&\end{aligned}}

Area $DGBHFD$

{\begin{aligned}=&\ 8Rmmpp+4Rpp-{\frac {16Rmmpp+4Rpp}{3}}\\\\=&\ {\frac {24Rmmpp+12Rpp-16Rmmpp-4Rpp}{3}}\\\\=&\ {\frac {8Rmmpp+8Rpp}{3}}\\\\=&\ {\frac {8Rp^{2}(m^{2}+1)}{3}}\end{aligned}}

Similarly it can be shown that Area $DGHD={\frac {4Rp^{2}(m^{2}+1)}{3}}$

Therefore line $AGH$  splits area $DGBHFD$  into two halves equal by area.

## Reflectivity of the Parabola

Figure 1: The Parabola $y={\frac {x^{2}}{4p}}$
Focus at point $F\ (0,p)$
Vertex at origin $(0,0)$
Directrix is line $y=-p$
By definition $BF=BH$  and $DF=DG$
In Figure 1 $p=1$

See Figure 1. $\triangle GFH$  is a right triangle and point $A$  is the midpoint of line $GAH$ .

$\therefore \ AG=AF=AH,\ \triangle ABF$  is congruent with $\triangle ABH$ , and $\angle ABF=\angle ABH$ .

$\angle CBJ=\angle ABH,\ \therefore \angle ABF=\angle CBJ$ .

1. Any ray of light emanating from a point source at F touches the parabola at B and is reflected away from B on a line that is always perpendicular to the directrix. $\angle ABF$  is the angle of incidence and $\angle CBJ$  is the angle of reflection.

2. The path from $K$  through focus to vertex and back to focus has length $JH$ . The path from $J$  to $B$  to $F$  has length $JH.\ \therefore$  all paths to and from focus have the same length.

In Theory

1) The quadratic may be used to examine itself.

Let a quadratic equation be: $y=x^{2}$ .

Let the equation of a line be: $y=mx+c$ .

Let the line intersect the quadratic at $(x_{1},y_{1})$ .

Therefore:

{\begin{aligned}y_{1}=mx_{1}+c\\c=y_{1}-mx_{1}\\y=x^{2}=mx+c=mx+(y_{1}-mx_{1})\\x^{2}-mx-y_{1}+mx_{1}=0\end{aligned}}

Let the line intersect the curve in exactly one place. Therefore $x$  must have exactly one value and the discriminant is $0$ .

{\begin{aligned}m^{2}-4(mx_{1}-y_{1})=0\\m^{2}-4mx_{1}+4y_{1}=0\\m^{2}-4mx_{1}+4x_{1}^{2}=0\\(m-2x_{1})^{2}=0\\m=2x_{1}\end{aligned}}

A line that touches the curve at $x_{1},y_{1}$  has slope $2x_{1}$ .

Therefore the slope of the curve at $x_{1},y_{1}$  is $2x_{1}$ . This examination of the curve has produced the slope of the curve without using calculus.

Consider the curve: $y=Ax^{2}+Bx+C$ . The aim is to calculate the slope of the curve at an arbitrary point $(x_{1},y_{1})$ .

{\begin{aligned}Ax^{2}+Bx+C=mx+c\\Ax^{2}+Bx+C=mx+(y_{1}-mx_{1})\\Ax^{2}+Bx+C-mx-y_{1}+mx_{1}=0\\Ax^{2}+Bx-mx+C-y_{1}+mx_{1}=0\\Ax^{2}+(B-m)x+(C-y_{1}+mx_{1})=0\end{aligned}}

If $x$  is to have exactly one value, discriminant $=(B-m)^{2}-4(A)(C-y_{1}+mx_{1})=0$ .

Therefore $BB-2Bm+mm-4AC+4Ay_{1}-4Amx_{1}=0$

$mm-4Ax_{1}m-2Bm+BB-4AC+4Ay_{1}=0$

$mm-(4Ax_{1}+2B)m+(BB-4AC+4Ay_{1})=0$

$m={\frac {(4Ax_{1}+2B)\pm {\sqrt {(4Ax_{1}+2B)^{2}-4(BB-4AC+4Ay_{1})}}}{2}}$

$m={\frac {(4Ax_{1}+2B)\pm {\sqrt {16AAx_{1}x_{1}+16ABx_{1}+4BB-4BB+16AC-16Ay_{1}}}}{2}}$

$m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {AAx_{1}x_{1}+ABx_{1}+AC-Ay_{1}}}}{2}}$

$m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {A(Ax_{1}^{2}+Bx_{1}+C-y_{1})}}}{2}}$

$m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {A(0)}}}{2}}$

$m=2Ax_{1}+B.$

The slope of the curve at an arbitrary point $(x_{1},y_{1})=2Ax_{1}+B$ .

2) The quadratic may be used to examine other curves, for example, the circle.

Define a circle of radius 5 at the origin:

${\sqrt {x^{2}+y^{2}}}=5$

$x^{2}+y^{2}=25$

Move the circle to $(8,3)$

$(x-8)^{2}+(y-3)^{2}=25$

$x^{2}-16x+64+y^{2}-6y+9=25$

$x^{2}-16x+y^{2}-6y+48=0$

We want to know the values of $x$  that contain the circle, that is, the values of $x$  for each of which there is only one value of $y$ .

Put the equation of the circle into a quadratic in $y$ .

$y^{2}-6y+x^{2}-16x+48=0$

$A=1,\ B=-6,\ C=x^{2}-16x+48$

There is exactly one value of $y$  if the discriminant is $0$ . Therefore

$(-6)(-6)-4(x^{2}-16x+48)=0$

$36-4(x^{2}-16x+48)=0$

$-9+x^{2}-16x+48=0$

$x^{2}-16x+39=0$

$(x-3)(x-13)=0$

$x_{1}=3,\ x_{2}=13$

These values of $x$  make sense because we expect the values of $x$  to be $8\pm 5$ . This process has calculated a minimum point and a maximum point without calculus.

3) The formula remains valid for $B$  and/or $C$  equal to $0$ . Under these conditions you probably won't need the formula. For example $Ax^{2}+Bx$  can be factored by inspection as $x(Ax+b)$ .

4) The quadratic can be used to solve functions of higher order.

One of the solutions of the cubic depends on the solution of a sextic in the form $Ax^{6}+Bx^{3}+C=0$ . This is the quadratic $AX^{2}+BX+C$  where $X=x^{3}$ .

The cubic function $x^{3}+6x^{2}+13x+10$  produces the depressed function $t^{3}+9t=t(t^{2}+9)$ .

The quadratic $t^{2}+9=0$  is solved as $t^{2}=-9,\ t={\sqrt {-9}}=\pm 3i$ .

The roots of the depressed function are $t_{1}=0,\ t_{2}=3i,\ t_{3}=-3i$ .

Using $x={\frac {-B+t}{3A}}$

$x_{1}={\frac {-6+0}{3}}=-2$

$x_{2}={\frac {-6+3i}{3}}=-2+i$

$x_{3}=-2-i$

In this example the quadratic, as part of the depressed function, simplified the solution of the cubic.

The quartic function $x^{4}+4x^{3}+17x^{2}+26x+11$  produces the depressed function $t^{4}+176t^{2}-256$  which is the quadratic $T^{2}+176T-256$  where $T=t^{2}$ .

5) The quadratic appears in Newton's Laws of Motion: $s=ut+{\frac {1}{2}}at^{2}$ Graph of quadratic function $y=x^{2}+4x-5$ showing basic features : * X and Y intercepts * vertex at (-2,-9), * axis of symmetry at x = -2. Graph of quadratic function $y=x^{2}-2x-3$ showing : * X and Y intercepts in red, * vertex and axis of symmetry in blue, * focus and directrix in pink.