The quadratic function has maximum power of ${\displaystyle x}$ equal to ${\displaystyle 2}$:

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

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

The solution of the quadratic equation is:

${\displaystyle x={\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\ \dots \ (2)}$

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

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

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

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

Substitute (3) into (1) and expand:

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

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

${\displaystyle (-B+t)(-B+t)+2B(-B+t)+4AC}$

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

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

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

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

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

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

p +- q

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

{\displaystyle {\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}}}

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

{\displaystyle {\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 ${\displaystyle (8)}$ into ${\displaystyle (10)}$

{\displaystyle {\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}}}

{\displaystyle {\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 ${\displaystyle f(x)=Ax^{2}+Bx+C}$ you suppose that ${\displaystyle f(p+q)=f(p-q)}$.

{\displaystyle {\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 ${\displaystyle (1)}$ is ${\displaystyle 2Ax+B}$ which equals ${\displaystyle 0}$ at a stationary point.

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

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

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

Substitute (12) in (1) and expand:

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

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

Substitute (13) in (1) and expand

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

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

If the function equals 0, then

{\displaystyle {\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).

p + qi

Let ${\displaystyle x=p+qi}$ where ${\displaystyle i={\sqrt {-1}}}$

Substitute this value of ${\displaystyle x}$ into (1) and expand:

{\displaystyle {\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 ${\displaystyle i=2Apqi+Bqi\ \dots \ (14)}$

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

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

From (15) and (16):

{\displaystyle {\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 ${\displaystyle i}$ 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 ${\displaystyle (-5,40),\ (4,13),\ (7,76)}$ are given and the three points satisfy ${\displaystyle y=f(x)}$, the values ${\displaystyle A,B,C}$ may be calculated.

{\displaystyle {\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 ${\displaystyle (17),\ (18),\ (19)}$ gives the equation ${\displaystyle y=2x^{2}-x-15}$.

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

By two points and a slope

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

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

Slope = 2Ax + B, therefore

{\displaystyle {\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 ${\displaystyle (20),(21),(22)}$ gives the polynomial ${\displaystyle 5x^{2}+22x+21\ \dots \ (23)}$.

By movement of the vertex

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

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

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

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

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

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

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

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

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

Therefore

{\displaystyle {\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 ${\displaystyle (23)}$ may be expressed as

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

For proof, expand:

{\displaystyle {\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: ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}$ (See [The General Quadratic] below.)

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

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

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

To express a valid quadratic in this way, both ${\displaystyle A,E}$ or both ${\displaystyle C,D}$ 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.

## The Parabola

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

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

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

${\displaystyle y={\frac {x^{2}}{4p}}}$ and slope = ${\displaystyle {\frac {x}{2p}}}$.

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

{\displaystyle {\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

${\displaystyle m}$ is the slope of line DB in Figure 1.

{\displaystyle {\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 ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (x_{2},y_{2})}$ are perpendicular.

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

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

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

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

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

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

Check our work:

{\displaystyle {\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 ${\displaystyle (2mp,-p)}$. They intersect on the directrix where ${\displaystyle y=-p}$. See [Tangents perpendicular and oscillating.]

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

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

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

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

The product of ${\displaystyle s_{3}}$ and ${\displaystyle s_{4}}$ is ${\displaystyle -1}$. Therefore the two sides are perpendicular and the triangle ${\displaystyle HGF}$ in Figure 1 is a right triangle.

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

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

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

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

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

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

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

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

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

Prove that tangent1 and tangent2 are perpendicular.

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

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

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

{\displaystyle {\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, ${\displaystyle m}$ is the slope of line ${\displaystyle DFB}$. This statement agrees with ${\displaystyle k=2mp}$ proved in the last section.)

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

Using ${\displaystyle y=mx+c}$,

{\displaystyle {\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 ${\displaystyle (x_{1},y_{1}),\ (x_{2},y_{2})}$ passes through the focus ${\displaystyle (0,p)}$.

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

Two lines parallel

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

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

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

Therefore two lines ${\displaystyle JGK,DFB}$ are parallel.

Area DGBHFD

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

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

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

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

Area under line ${\displaystyle DFB}$

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

Area under curve ${\displaystyle DGB}$

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

Area ${\displaystyle DGBHFD}$

{\displaystyle {\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 ${\displaystyle DGHD={\frac {4Rp^{2}(m^{2}+1)}{3}}}$

Therefore line ${\displaystyle AGH}$ splits area ${\displaystyle DGBHFD}$ into two halves equal by area.

## Reflectivity of the Parabola

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

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

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

${\displaystyle \angle CBJ=\angle ABH,\ \therefore \angle ABF=\angle CBJ}$.

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. ${\displaystyle \angle ABF}$ is the angle of incidence and ${\displaystyle \angle CBJ}$ is the angle of reflection.

2. The path from ${\displaystyle K}$ through focus to vertex and back to focus has length ${\displaystyle JH}$. The path from ${\displaystyle J}$ to ${\displaystyle B}$ to ${\displaystyle F}$ has length ${\displaystyle JH.\ \therefore }$ 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: ${\displaystyle y=x^{2}}$.

Let the equation of a line be: ${\displaystyle y=mx+c}$.

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

Therefore:

{\displaystyle {\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 ${\displaystyle x}$ must have exactly one value and the discriminant is ${\displaystyle 0}$.

{\displaystyle {\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 ${\displaystyle x_{1},y_{1}}$ has slope ${\displaystyle 2x_{1}}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Define a circle of radius 5 at the origin:

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

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

Move the circle to ${\displaystyle (8,3)}$

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

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

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

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

Put the equation of the circle into a quadratic in ${\displaystyle y}$.

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

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

There is exactly one value of ${\displaystyle y}$ if the discriminant is ${\displaystyle 0}$. Therefore

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

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

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

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

${\displaystyle (x-3)(x-13)=0}$

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

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

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

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 ${\displaystyle Ax^{6}+Bx^{3}+C=0}$. This is the quadratic ${\displaystyle AX^{2}+BX+C}$ where ${\displaystyle X=x^{3}}$.

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

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

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

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

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

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

${\displaystyle x_{3}=-2-i}$

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

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

5) The quadratic appears in Newton's Laws of Motion: ${\displaystyle s=ut+{\frac {1}{2}}at^{2}}$

Let the directrix be ${\displaystyle ax+by+c=0}$ where at least one of ${\displaystyle a,b}$ is non-zero.

Let the focus be ${\displaystyle (p,q)}$.

Let ${\displaystyle (x,y)}$ be any point on the curve.

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

Distance from point ${\displaystyle (x,y)}$ to directrix (${\displaystyle ax+by+c=0}$)

= ${\displaystyle {\frac {{\sqrt {R}}(ax+by+c)}{R}}}$ where ${\displaystyle R=a^{2}+b^{2}}$.

By definition these two lengths are equal:

${\displaystyle {\sqrt {(x-p)^{2}+(y-q)^{2}}}={\frac {{\sqrt {R}}(ax+by+c)}{R}}}$.

${\displaystyle \therefore \ R{\sqrt {(x-p)^{2}+(y-q)^{2}}}={\sqrt {R}}(ax+by+c)}$.

Square both sides:

${\displaystyle R^{2}((x-p)^{2}+(y-q)^{2})=R(ax+by+c)^{2}}$.

${\displaystyle R((x-p)^{2}+(y-q)^{2})=(ax+by+c)^{2}}$.

${\displaystyle R((x-p)^{2}+(y-q)^{2})-(ax+by+c)^{2}=0}$.

Expand and the result is:

${\displaystyle b^{2}x^{2}-2abxy+a^{2}y^{2}-2(ac+pR)x-2(bc+qR)y+R(p^{2}+q^{2})-c^{2}=0\ \dots \ (25)}$.

${\displaystyle (25)}$ has the form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ where

{\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}}}

If ${\displaystyle a==0}$ then

${\displaystyle b^{2}x^{2}-2(pR)x-2(bc+qR)y+R(p^{2}+q^{2})-c^{2}=0}$.

${\displaystyle R=b^{2}}$

${\displaystyle 2(bc+qb^{2})y=b^{2}x^{2}-2(pb^{2})x+b^{2}(p^{2}+q^{2})-c^{2}}$.

${\displaystyle 2b(c+qb)y=b^{2}x^{2}-2pb^{2}x+b^{2}(p^{2}+q^{2})-c^{2}\ \dots \ (26)}$.

Let

{\displaystyle {\begin{aligned}b=1\\c=1\\p=0\\q=1\end{aligned}}}

Directrix has equation: ${\displaystyle (0)x+(1)y+(1)=0}$ or ${\displaystyle y=-1}$.

Focus has coordinates ${\displaystyle (0,1)}$.

${\displaystyle (26)}$ becomes ${\displaystyle y={\frac {x^{2}}{4}}}$. See Figure 1.

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

Let

{\displaystyle {\begin{aligned}b=1\\c=3\\p=4\\q=1\end{aligned}}}

Directrix has equation: ${\displaystyle (0)x+(1)y+(3)=0}$ or ${\displaystyle y=-3}$.

Focus has coordinates ${\displaystyle (4,1)}$.

${\displaystyle (26)}$ becomes ${\displaystyle 8y=x^{2}-8x+8}$ or ${\displaystyle 8(y+1)=(x-4)^{2}}$. See Figure 3.

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

Let

{\displaystyle {\begin{aligned}a=1\\b=0\\c=3\\p=1\\q=4\end{aligned}}}

Directrix has equation: ${\displaystyle (1)x+(0)y+(3)=0}$ or ${\displaystyle x=-3}$.

Focus has coordinates ${\displaystyle (1,4)}$.

${\displaystyle (25)}$ becomes ${\displaystyle 8x=y^{2}-8y+8}$ or ${\displaystyle 8(x+1)=(y-4)^{2}}$. See Figure 4.

An interesting situation occurs if the focus is on the directrix. Consider the directrix:

${\displaystyle 4x-3y+15=0}$

and the focus ${\displaystyle (3,9)}$ which is on the directrix.

${\displaystyle a=4,\ b=-3,\ c=15,\ p=3,\ q=9}$

${\displaystyle (25)}$ becomes:

${\displaystyle 9xx+24xy+16yy-270x-360y+2025=0}$

This seems to be the equation of a parabola because ${\displaystyle B^{2}-4AC=0}$, but look closely.

${\displaystyle 9xx+24xy+16yy-270x-360y+2025=(3x+4y-45)^{2}}$

The result is a line through the focus and normal to the directrix.

## Reverse-Engineering the Parabola

Figure 5: Parabola with 2 tangents parallel to axes.
Tangent ${\displaystyle ABC:y={\frac {16}{3}}}$.
Tangent ${\displaystyle ADE:x=-0.25}$.
Point A on directrix.
Line ${\displaystyle AG}$ perpendicular to line ${\displaystyle BD}$
Focus at ${\displaystyle F}$.

Given a parabola in form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ the aim is to produce the directrix and the focus.

We will solve the example shown in Figure 5: ${\displaystyle 9x^{2}-24xy+16y^{2}+70x-260y+1025=0}$,

where:

{\displaystyle {\begin{aligned}A&=9\\B&=-24\\C&=16\\D&=70\\E&=-260\\F&=1025\end{aligned}}}

${\displaystyle a={\sqrt {C}}=4;\ b={\frac {-B}{2a}}={\frac {-(-24)}{8}}=3}$.

Method 1. By analytical geometry

Find two tangents that intersect at a right angle. The simplest to find are those that are parallel to the axes.

Put the equation of the parabola in the form of a quadratic in ${\displaystyle x}$.

${\displaystyle Ax^{2}+Bxy+Dx+Cy^{2}+Ey+F=0}$

${\displaystyle Ax^{2}+(By+D)x+(Cy^{2}+Ey+F)=0}$

At the tangent there is exactly one value of ${\displaystyle x}$. Therefore the discriminant must be ${\displaystyle 0}$.

${\displaystyle (By+D)^{2}-4(A)(Cy^{2}+Ey+F)=0}$

${\displaystyle BByy+2BDy+DD-4ACyy-4AEy-4AF=0}$

${\displaystyle (BB-4AC)yy+(2BD-4AE)y+(DD-4AF)=0}$

In the general parabola ${\displaystyle B^{2}-4AC=0}$ therefore ${\displaystyle y={\frac {4AF-DD}{2BD-4AE}}}$.

In this example ${\displaystyle y={\frac {16}{3}};\ x={\frac {-(By+D)}{2A}}={\frac {29}{9}}}$.

Point ${\displaystyle B(x_{1},y_{1})}$ has coordinates ${\displaystyle ({\frac {29}{9}},\ {\frac {16}{3}})}$.

The line ${\displaystyle ABC}$ is tangent to the curve at ${\displaystyle B}$ and has equation: ${\displaystyle y={\frac {16}{3}}}$.

Put the equation of the parabola in the form of a quadratic in ${\displaystyle y}$:

${\displaystyle Cy^{2}+(Bx+E)y+(Ax^{2}+Dx+F)=0}$.

By using calculations similar to the above, ${\displaystyle x={\frac {4CF-EE}{2BE-4CD}}=-0.25}$ and ${\displaystyle y={\frac {-(Bx+E)}{2C}}=7.9375}$.

Point ${\displaystyle D(x_{2},y_{2})}$ has coordinates ${\displaystyle (-0.25,\ 7.9375)}$.

The line ${\displaystyle ADE}$ is tangent to the curve at ${\displaystyle D}$ and has equation: ${\displaystyle x=-0.25}$.

Point ${\displaystyle A}$ at the intersection of the two tangents has coordinates ${\displaystyle (x_{2},y_{1})=(-0.25,\ {\frac {16}{3}})}$, and point ${\displaystyle A}$ is on the directrix, the equation of which is: ${\displaystyle ax+by+c=0}$.

Put known values into the equation of the directrix: ${\displaystyle (4)(-0.25)+(3){\frac {16}{3}}+c=0}$.

Therefore ${\displaystyle c=-15}$ and the equation of the directrix is: ${\displaystyle 4x+3y-15=0}$.

The coordinates of points ${\displaystyle B,D}$ are known. Therefore line ${\displaystyle BD}$ is defined as: ${\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y-6.2=0}$.

Draw the line ${\displaystyle AG}$ perpendicular to ${\displaystyle BD}$. The line ${\displaystyle AG}$ is defined as: ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+3.4=0}$.

Point ${\displaystyle F}$ at the intersection of lines ${\displaystyle BD,AG}$ is the focus with coordinates ${\displaystyle (1,7)}$.

Method 2. By algebra

{\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: ${\displaystyle p,q,c}$:

{\displaystyle {\begin{aligned}D+2ac+2pR=0\\E+2bc+2qR=0\\F-Rpp-Rqq+cc=0\end{aligned}}}

The solutions are:

{\displaystyle {\begin{aligned}p&={\frac {D^{2}R+2DEab+4FRa^{2}-(Db)^{2}-(Ea)^{2}}{4R(Db^{2}-Eab-DR)}}\\q&={\frac {Db-Ea+2Rbp}{2Ra}}\\c&={\frac {-(D+2Rp)}{2a}}\end{aligned}}}

If ${\displaystyle a}$ is ${\displaystyle 0}$, the following may be used:

{\displaystyle {\begin{aligned}p&={\frac {-D}{2Rb}}\\q&={\frac {D^{2}-E^{2}-4FR}{4ER}}\\c&={\frac {-(E+2Rq)}{2b}}\end{aligned}}}

## Area enclosed between parabola and chord

Figure 6: The Parabola: ${\displaystyle y=x^{2}}$
Chord ${\displaystyle DC}$, parallel tangent ${\displaystyle AB}$ and area ${\displaystyle DOCD}$.
Chord ${\displaystyle OC}$, parallel tangent ${\displaystyle GHI}$ and area ${\displaystyle OHCO}$.

See Figure 6. The curve is: ${\displaystyle y=x^{2}}$. Integral is: ${\displaystyle {\frac {x^{3}}{3}}}$.

Area under curve ${\displaystyle (OBC)}$

{\displaystyle {\begin{aligned}x=&1\\=\ \ \ \ \ &[{\frac {x^{3}}{3}}]={\frac {1}{3}}\\x=&0\end{aligned}}}

Area under curve ${\displaystyle (DOC)=}$ area${\displaystyle (OAD)+}$ area${\displaystyle (OBC)={\frac {2}{3}}}$

Area between chord ${\displaystyle DC}$ and curve ${\displaystyle DOC=2-{\frac {2}{3}}={\frac {4}{3}}}$.

Consider the chord ${\displaystyle CD}$. Call this the ${\displaystyle base}$ with value ${\displaystyle 2}$. The tangent ${\displaystyle AOB}$ through the origin is parallel to ${\displaystyle base\ (DC)}$, and the perpendicular distance between ${\displaystyle AB,DC}$ is ${\displaystyle 1}$. Call this distance the ${\displaystyle height}$ with value ${\displaystyle 1}$.

In this case the area enclosed between chord ${\displaystyle DC}$ and curve ${\displaystyle DOC={\frac {2}{3}}(base)(height)={\frac {2}{3}}(2)(1)={\frac {4}{3}},}$ the same as that calculated earlier.

Consider the chord ${\displaystyle OC}$ and curve ${\displaystyle OHC.}$ By inspection the area between chord ${\displaystyle OC}$ and curve ${\displaystyle OHC={\frac {1}{2}}-{\frac {1}{3}}={\frac {1}{6}}.}$

Chord ${\displaystyle OC}$ has equation ${\displaystyle y=x;\ x-y=0;\ {\frac {x}{\sqrt {2}}}-{\frac {y}{\sqrt {2}}}=0}$ in normal form.

The line ${\displaystyle GHI}$ is parallel to ${\displaystyle base\ OC}$ and touches the curve at ${\displaystyle H({\frac {1}{2}},\ {\frac {1}{4}}).}$

Distance from ${\displaystyle H}$ to chord ${\displaystyle OC={\frac {1/2}{\sqrt {2}}}-{\frac {1/4}{\sqrt {2}}}={\frac {1}{4{\sqrt {2}}}}=height.}$

Length of ${\displaystyle OC={\sqrt {2}}=base.}$

Area between chord ${\displaystyle OC}$ and curve ${\displaystyle OHC={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\sqrt {2}})({\frac {1}{4{\sqrt {2}}}})={\frac {2}{3}}({\frac {1}{4}})={\frac {1}{6}},}$ the same as that calculated earlier.

These observations suggest that the area enclosed between chord and curve ${\displaystyle ={\frac {2}{3}}(base)(height).}$

Figure 7: The Parabola: ${\displaystyle y=Kx^{2}}$
Chord ${\displaystyle IJ}$, parallel tangent ${\displaystyle FG}$ and area ${\displaystyle IOJI}$, the area enclosed between chord ${\displaystyle IJ}$ and curve.

We prove this identity for the general case. See Figure 7.

Slope of chord ${\displaystyle IJ={\frac {Kqq-Kpp}{q-p}}={\frac {K((q+p)(q-p))}{q-p}}=K(q+p).}$

Find equation of chord ${\displaystyle IJ.\ y=K(p+q)x+c;\ \therefore c=y-K(p+q)x=Kqq-K(p+q)q=-Kpq}$

Equation of chord ${\displaystyle IJ:\ y=K(p+q)x-Kpq.}$

Find equation of tangent ${\displaystyle FG.}$

{\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 ${\displaystyle c}$ that gives ${\displaystyle x}$ exactly one value.

Therefore discriminant ${\displaystyle K^{2}(p+q)^{2}+4Kc=0;\ c={\frac {-K(p+q)^{2}}{4}}.}$

${\displaystyle y=K(p+q)x-{\frac {K(p+q)^{2}}{4}};\ K(p+q)x-y-{\frac {K(p+q)^{2}}{4}}=0;}$

Equation of tangent ${\displaystyle FG}$ in normal form: ${\displaystyle {\frac {K(p+q)x-y-{\frac {K(p+q)^{2}}{4}}}{\sqrt {K^{2}(p+q)^{2}+1}}}=0.}$

Equation of chord ${\displaystyle IJ}$ in normal form: ${\displaystyle {\frac {K(p+q)x-y-Kpq}{\sqrt {K^{2}(p+q)^{2}+1}}}=0.}$

Therefore distance between chord ${\displaystyle IJ}$ and tangent ${\displaystyle FG}$

${\displaystyle =height={\frac {-Kpq-(-{\frac {K(p+q)^{2}}{4}})}{R}}={\frac {K(p+q)^{2}-4Kpq}{4R}}={\frac {K(p-q)^{2}}{4R}}}$ where ${\displaystyle R={\sqrt {K^{2}(p+q)^{2}+1}}.}$

Length of chord ${\displaystyle IJ=base=L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}.}$

Area under chord ${\displaystyle IJ=(q-p){\frac {Kqq+Kpp}{2}}={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}}$

Area under curve ${\displaystyle IOJ}$

{\displaystyle {\begin{aligned}x=&q\\=\ \ \ \ \ &[{\frac {Kx^{3}}{3}}]={\frac {Kqqq-Kppp}{3}}\\x=&p\end{aligned}}}

Area between chord ${\displaystyle IJ}$ and curve ${\displaystyle IOJ}$

${\displaystyle ={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}-{\frac {Kqqq-Kppp}{3}}={\frac {Kqqq+3Kppq-3Kpqq-Kppp}{6}}={\frac {S}{6}}.}$

The aim is to prove that:

${\displaystyle {\frac {2}{3}}(base)(height)={\frac {S}{6}}}$ or

${\displaystyle {\frac {2}{3}}(L)({\frac {K(p-q)^{2}}{4R}})={\frac {S}{6}}}$ or

${\displaystyle {\frac {2LK(p-q)^{2}}{12R}}={\frac {S}{6}}}$ or

${\displaystyle LK(p-q)^{2}=RS}$

where:

${\displaystyle L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}}$

${\displaystyle R={\sqrt {K^{2}(p+q)^{2}+1}}}$

${\displaystyle S=Kqqq+3Kppq-3Kpqq-Kppp}$

Because ${\displaystyle L,R}$ contain a square root, we prove that the squares are equal:

${\displaystyle L^{2}K^{2}((p-q)^{2})^{2}=R^{2}S^{2}}$

where:

${\displaystyle L^{2}=(Kqq-Kpp)^{2}+(q-p)^{2};\ R^{2}=K^{2}(p+q)^{2}+1}$

If you choose to make the substitutions and go through all the calculations, you will find that the squares are equal.

Therefore |${\displaystyle LK(p-q)^{2}}$| = |${\displaystyle RS}$| and area enclosed between curve and chord =${\displaystyle {\frac {2}{3}}(base)(height)}$

where ${\displaystyle base}$ is the length of the chord, and ${\displaystyle height}$ is the perpendicular distance between chord and tangent parallel to chord.

## Gallery

 Graph of quadratic function ${\displaystyle y=x^{2}+4x-5}$ showing basic features : * X and Y intercepts * vertex at (-2,-9), * axis of symmetry at x = -2. Graph of quadratic function ${\displaystyle y=x^{2}-2x-3}$ showing : * X and Y intercepts in red, * vertex and axis of symmetry in blue, * focus and directrix in pink.