# Parabola

Figure 1: The Parabola ${\displaystyle y={\frac {x^{2}}{4q}}}$
Focus at point ${\displaystyle F\ (0,q)}$
Vertex at origin ${\displaystyle (0,0)}$
Directrix is line ${\displaystyle y=-q}$
By definition ${\displaystyle P_{1}F=P_{1}D_{1},\ P_{2}F=P_{2}D_{2},\ P_{4}F=P_{4}D_{4}.}$
${\displaystyle VF=VD_{3}=q.}$
In Figure 1 ${\displaystyle q=1}$

In Cartesian geometry in two dimensions the ${\displaystyle parabola}$ 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 ${\displaystyle focus}$ and the fixed line is called the ${\displaystyle directrix}$. Distance from ${\displaystyle focus}$ to ${\displaystyle directrix}$ is non-zero.

See Figure 1.

The focus is point ${\displaystyle F\ (0,q)}$ and the directrix is line ${\displaystyle D_{1}D_{4}}$${\displaystyle :\ y=-q.}$ The ${\displaystyle vertex}$, point ${\displaystyle V}$${\displaystyle :\ (0,0)}$, is half-way between focus and directrix. A ${\displaystyle chord}$ is the segment of a line joining any two distinct points of the parabola. The line segment ${\displaystyle P_{2}FP_{4}}$ is a chord. Because chord ${\displaystyle P_{2}FP_{4}}$ passes through the focus ${\displaystyle F}$, it is called a ${\displaystyle focal\ chord.}$

The focal chord parallel to the directrix is called the ${\displaystyle latus\ rectum.}$

The line through the focus and perpendicular to the directrix is the ${\displaystyle axis}$, sometimes called ${\displaystyle axis\ of\ symmetry}$.

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

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

${\displaystyle x^{2}-4qy=0;\ y={\frac {x^{2}}{4q}}}$.

If the equation of the curve is expressed as: ${\displaystyle y=Kx^{2}}$, then ${\displaystyle K={\frac {1}{4q}};\ 4Kq=1;\ q={\frac {1}{4K}}}$ where the ${\displaystyle focus}$ has coordinates ${\displaystyle (0,q),}$ and ${\displaystyle q}$ is the distance form vertex to focus.

If the directrix is parallel to the ${\displaystyle X}$ 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

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 (${\displaystyle ax+by+c=0}$)

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

By definition these two lengths are equal:

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

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

Square both sides:

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

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

Expand and the result is:

${\displaystyle b^{2}x^{2}-2abxy+a^{2}y^{2}-2(ac+pR)x-2(bc+qR)y+R(p^{2}+q^{2})-c^{2}=0\ \dots \ (1)}$.

${\displaystyle (1)}$ has the form of the equation of the conic section ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ 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}}}

${\displaystyle B^{2}=4AC}$ because this curve is a parabola.

## An Example

Figure 2: The Parabola ${\displaystyle 16x^{2}+24xy+9y^{2}-20x-140y+100=0}$
Green line is ${\displaystyle directrix:\ 3x-4y-5=0}$
Blue line is ${\displaystyle axis:\ 4x+3y-10=0}$
Focus at point ${\displaystyle F:\ (1,2)}$
Vertex at point ${\displaystyle V:\ (1.6,1.2)}$
${\displaystyle VF=1.}$ Shape of curve is: ${\displaystyle y={\frac {x^{2}}{4}}.}$

See Figure 2.

${\displaystyle (a,b,c)=(3,-4,-5)}$

${\displaystyle (p,q)=(1,2)}$

The equation of the parabola is derived as follows:

{\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: ${\displaystyle 16x^{2}+24xy+9y^{2}-20x-140y+100=0.}$

Equation of directrix in normal form: ${\displaystyle {\frac {3}{5}}x-{\frac {4}{5}}y-1=0.}$

Distance from ${\displaystyle focus}$ to ${\displaystyle directrix={\frac {3}{5}}(1)-{\frac {4}{5}}(2)-1=-2.}$

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

Therefore, curve has shape of ${\displaystyle y=Kx^{2}}$ where ${\displaystyle K={\frac {1}{4}}}$.

 Caution: 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. ${\displaystyle a=4,\ b=-3,\ c=15,\ p=3,\ q=9}$ In this case the "parabola" has equation: ${\displaystyle 9xx+24xy+16yy-270x-360y+2025=0}$. 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. If you solve for ${\displaystyle p,q,c}$ using the algebraic solutions, you will produce the values ${\displaystyle 3,9,15}$ as above. However, the distance between focus and directrix = ${\displaystyle {\frac {4}{5}}x+{\frac {-3}{5}}y+3}$ where ${\displaystyle x=p;\ y=q;}$ distance ${\displaystyle ={\frac {4}{5}}(3)+{\frac {-3}{5}}(9)+3=0.}$

## Reverse-Engineering the Parabola

Figure 3: Parabola with 2 tangents parallel to axes.
Tangent ${\displaystyle ABC:y={\frac {16}{3}}}$.
Tangent ${\displaystyle ADE:x=-0.25}$.
Point ${\displaystyle A}$ on directrix, oblique, thin, black line.${\displaystyle }$
Line ${\displaystyle AFG}$ perpendicular to focal chord ${\displaystyle BFD.}$
Focus at ${\displaystyle F:(1,7)}$.

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 3: ${\displaystyle 9x^{2}-24xy+16y^{2}+70x-260y+1025=0}$,

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 chord ${\displaystyle BD}$ 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:

${\displaystyle c={\frac {4FR-(DD+EE)}{4Da+4Eb}}}$

${\displaystyle p={\frac {-(D+2ac)}{2R}}}$

${\displaystyle q={\frac {-(E+2bc)}{2R}}}$

 If ${\displaystyle b}$ is ${\displaystyle 0,\ A=B=0,\ R=a^{2}=C,}$ the parabola becomes the quadratic: ${\displaystyle -Dx=Cy^{2}+Ey+F}$ and: {\displaystyle {\begin{aligned}p&={\frac {E^{2}-D^{2}-4FC}{4CD}}\\q&={\frac {-E}{2C}}\\c&={\frac {4FC-(DD+EE)}{4Da}}\end{aligned}}} The directrix has equation: ${\displaystyle ax+c=0;\ ax=-c;\ x={\frac {DD+EE-4FC}{4CD}}}$. If ${\displaystyle D}$ is ${\displaystyle -1}$, then: {\displaystyle {\begin{aligned}x&=Cy^{2}+Ey+F\\p&={\frac {E^{2}-1-4FC}{4C(-1)}}={\frac {1-(E^{2}-4CF)}{4C}}\\q&={\frac {-E}{2C}}\end{aligned}}} The directrix has equation: ${\displaystyle x={\frac {-1-(E^{2}-4CF)}{4C}}}$.
 If ${\displaystyle a}$ is ${\displaystyle 0,\ B=C=0,\ R=b^{2}=A,}$ the parabola becomes the quadratic: ${\displaystyle -Ey=Ax^{2}+Dx+F}$ and: {\displaystyle {\begin{aligned}p&={\frac {-D}{2A}}\\q&={\frac {D^{2}-E^{2}-4FA}{4EA}}\\c&={\frac {4FA-(DD+EE)}{4Eb}}\end{aligned}}} The directrix has equation: ${\displaystyle by+c=0;\ by=-c;\ y={\frac {DD+EE-4FA}{4EA}}}$. 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. If ${\displaystyle E}$ is ${\displaystyle -1}$, then: {\displaystyle {\begin{aligned}y&=Ax^{2}+Dx+F\\p&={\frac {-D}{2A}}\\q&={\frac {D^{2}-1-4FA}{4(-1)A}}={\frac {1-(D^{2}-4AF)}{4A}}\end{aligned}}} The directrix has equation: ${\displaystyle y={\frac {-1-(D^{2}-4AF)}{4A}}}$. These values agree with the corresponding values in the graph of ${\displaystyle y=x^{2}-2x-3}$ to the right.

## Slope of the Parabola

Consider parabola ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ and line ${\displaystyle y=mx+c.}$

Let point ${\displaystyle (X,Y)}$ be any point on the line. Therefore ${\displaystyle Y=mX+c;\ c=Y-mX;\ y=mx+(Y-mX).}$

Let the line intersect the parabola. Substitute the above value of ${\displaystyle y}$ into the equation of the parabola and expand:

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

Discriminant =

{\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 ${\displaystyle m}$:

{\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}}}

${\displaystyle m={\frac {-(+8ACXY+4AEX-2BBXY-2BDX-2BEY-4BF+4CDY+2DE)\pm {\sqrt {R}}}{2(-4ACXX+BBXX+2BEX-4CDX-4CF+EE)}}}$

where:

{\displaystyle {\begin{aligned}R=&\ 16(AXX+BXY+CYY+DX+EY+F)(-4ACF+AEE+BBF-BDE+CDD)\end{aligned}}}

If the point ${\displaystyle (X,Y)}$ is on the curve, then the line touches the curve at ${\displaystyle (X,Y)}$ and:

{\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 ${\displaystyle B^{2}-4AC=0}$ for a parabola.

When slope is displayed in this format, we see that slope is vertical if ${\displaystyle (2BE-4CD)X+EE-4CF=0.}$

The line ${\displaystyle x={\frac {4CF-EE}{2BE-4CD}}}$ is tangent to the curve.

 Let the equation of a line be: ${\displaystyle x=My+c}$ in which case ${\displaystyle M={\frac {1}{m}}.}$ By using calculations similar to the above it can be shown that: ${\displaystyle M={\frac {(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(2BD-4AE)Y+DD-4AF}}}$. ${\displaystyle m={\frac {1}{M}}}$ therefore: ${\displaystyle m={\frac {(2BD-4AE)Y+DD-4AF}{(BD-2AE)X+(BE-2CD)Y+2BF-DE}}}$ When slope is displayed in this format, we see that slope is zero if ${\displaystyle (2BD-4AE)Y+DD-4AF=0}$. The line ${\displaystyle y={\frac {4AF-DD}{2BD-4AE}}}$ is tangent to the curve.
 This examination of the parabola has produced two expressions for slope of the parabola: ${\displaystyle m={\frac {(2BD-4AE)y+DD-4AF}{(BD-2AE)x+(BE-2CD)y+2BF-DE}}={\frac {(BD-2AE)x+(BE-2CD)y+2BF-DE}{(2BE-4CD)x+EE-4CF}}}$. where the point ${\displaystyle (x,y)}$ is any arbitrary point on the curve. Therefore ${\displaystyle m=\pm {\sqrt {\frac {(2BD-4AE)y+DD-4AF}{(2BE-4CD)x+EE-4CF}}}}$. This formula for ${\displaystyle m}$ contains both tangents parallel to the axes and is derived without calculus.
 The formula from calculus below is simpler and unambiguous concerning sign. {\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 ${\displaystyle 0}$ where ${\displaystyle (2BD-4AE)y+DD-4AF=0;\ 2Ax+By+D=0;}$ or: ${\displaystyle y={\frac {4AF-DD}{2BD-4AE}};\ x={\frac {-(By+D)}{2A}}.}$ The slope of the parabola is vertical where ${\displaystyle (2BE-4CD)x+EE-4CF=0;\ Bx+2Cy+E=0;}$ or: ${\displaystyle x={\frac {4CF-EE}{2BE-4CD}};\ y={\frac {-(Bx+E)}{2C}}.}$
 Caution: If the curve is ${\displaystyle y=Kx^{2}}$, the slope can never be vertical. If the curve is ${\displaystyle x=Ky^{2}}$, the slope can never be ${\displaystyle 0}$.
 If ${\displaystyle B=C=0}$ and ${\displaystyle E=-1}$, the equation of the parabola becomes: ${\displaystyle y=Ax^{2}+Dx+F}$ and: {\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

Given parabola defined by ${\displaystyle A,B,C,D,E,F}$ and slope ${\displaystyle m={\frac {yvalue}{xvalue}}}$ where at least one of ${\displaystyle yvalue,xvalue}$ is non-zero, calculate point at which the slope is ${\displaystyle m.}$

Let ${\displaystyle m={\frac {-(2Ax+By+D)}{Bx+2Cy+E}}={\frac {s}{t}}={\frac {yvalue}{xvalue}}}$

Then ${\displaystyle s(Bx+2Cy+E)+t(2Ax+By+D)=0.}$

Let ${\displaystyle G=(Bs+2At);\ H=(2Cs+Bt);\ I=(Es+Dt).}$

Then ${\displaystyle Gx+Hy+I=0\ \dots \ (1)}$

Substitute in the equation of the parabola and ${\displaystyle y={\frac {-(AII-DGI+FGG)}{(2AHI-BGI-DGH+EGG)}}\ \dots \ (2)}$

As shown below, with a little manipulation of the data, the same formula can be used to calculate ${\displaystyle x.}$

 Equation ${\displaystyle (1)}$ above is the equation of a straight line with slope ${\displaystyle {\frac {-G}{H}}.}$ Substitute for ${\displaystyle G,H}$ and the slope of ${\displaystyle (1)}$ becomes ${\displaystyle {\frac {b}{a}}.}$ Equation ${\displaystyle (1)}$ is that of a line parallel to the axis of symmetry of the parabola. It is possible for both both ${\displaystyle G,H}$ to equal ${\displaystyle 0}$ in which case the calculation of ${\displaystyle y,(2)}$ above fails as an attempt to divide by ${\displaystyle 0.}$ See caution under "Slope of the Parabola" above.

#### Implementation

 # 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 
##### Examples

A parabola is defined as ${\displaystyle 9x^{2}-24xy+16y^{2}+70x-260y+1025=0.}$

 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)  (0.2, 6.4) 
 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)  (-0.25, 7.9375) 

## Parabola and any chord

Figure 4: Parabola and any chord.
Origin ${\displaystyle (0,0)}$ at point ${\displaystyle O;}$ curve${\displaystyle :}$ ${\displaystyle y=Kx^{2};}$
chord ${\displaystyle IJ}$${\displaystyle :\ y=mx+c;}$ point ${\displaystyle L}$${\displaystyle :\ (0,c);}$
2 tangents${\displaystyle :}$ ${\displaystyle IN,JN;}$ point ${\displaystyle N}$${\displaystyle :\ ({\frac {m}{2K}},-c)}$

Refer to Figure 4.

The curve has equation: ${\displaystyle y=Kx^{2}.}$

The chord ${\displaystyle IJ}$ has equation: ${\displaystyle y=mx+c}$.

Point ${\displaystyle L}$ has coordinates ${\displaystyle (0,c).}$

Line ${\displaystyle IN}$ is tangent to the curve at ${\displaystyle I.}$

Line ${\displaystyle JN}$ is tangent to the curve at ${\displaystyle J.}$

This section shows that point ${\displaystyle N}$ has coordinates ${\displaystyle ({\frac {m}{2K}},-c).}$

{\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 ${\displaystyle R={\sqrt {m^{2}+4Kc}}}$

Point ${\displaystyle I}$ has coordinates ${\displaystyle (x_{1},y_{1})}$ where:

${\displaystyle x_{1}={\frac {m-R}{2K}}}$,

${\displaystyle y_{1}=Kx_{1}^{2}=K({\frac {m-R}{2K}})({\frac {m-R}{2K}})={\frac {m^{2}-2mR+R^{2}}{4K}}={\frac {m^{2}-mR+2Kc}{2K}}}$,

and slope of tangent ${\displaystyle IN=s_{1}=2Kx_{1}=m-R.}$

Point ${\displaystyle J}$ has coordinates ${\displaystyle (x_{2},y_{2})}$ where:

${\displaystyle x_{2}={\frac {m+R}{2K}}}$,

${\displaystyle y_{2}={\frac {m^{2}+mR+2Kc}{2K}}}$,

and slope of tangent ${\displaystyle JN=s_{2}=m+R.}$

Points ${\displaystyle D,E}$ have coordinates ${\displaystyle (x_{1},0),(x_{2},0).}$

Equation of tangent ${\displaystyle IN:}$

{\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 ${\displaystyle JN:}$

{\displaystyle {\begin{aligned}y=(m+R)x-{\frac {m^{2}+mR+2Kc}{2K}}\end{aligned}}}

At point of intersection ${\displaystyle N,\ (m+R)x-{\frac {m^{2}+mR+2Kc}{2K}}=(m-R)x-{\frac {m^{2}-mR+2Kc}{2K}};\ x={\frac {m}{2K}}.}$

Review the ${\displaystyle X}$ coordinates of points ${\displaystyle D,E}$${\displaystyle :\ ({\frac {m-R}{2K}},{\frac {m+R}{2K}})}$.

The line ${\displaystyle NGH}$ with equation ${\displaystyle x={\frac {m}{2K}}}$ bisects the line segment ${\displaystyle DE}$ and also the chord ${\displaystyle IJ}$ at point ${\displaystyle H}$.

Any chord parallel to ${\displaystyle IJ}$ has two tangents that intersect on the line ${\displaystyle x={\frac {m}{2K}}}$.

Any chord parallel to ${\displaystyle IJ}$ is bisected by the line ${\displaystyle x={\frac {m}{2K}}}$.

The ${\displaystyle Y}$ coordinate of point ${\displaystyle N:}$

${\displaystyle }$

{\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 ${\displaystyle L\ (0,c)}$ has two tangents that intersect on the line ${\displaystyle y=-c}$.

Angle ${\displaystyle INJ}$

 Using: {\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 ${\displaystyle 4Kc>1}$, point ${\displaystyle L}$ is above the ${\displaystyle focus,\ \tan(\angle INJ)}$ is positive and ${\displaystyle 0}$° ${\displaystyle <\angle INJ<90}$°. ${\displaystyle \angle INJ}$ is acute and, as ${\displaystyle m}$ increases, ${\displaystyle \angle INJ}$ increases, approaching ${\displaystyle 90}$°. If ${\displaystyle 4Kc==1}$, point ${\displaystyle L}$ is on the ${\displaystyle focus,\ \tan(\angle INJ)={\frac {2{\sqrt {m^{2}+1}}}{0}},\ \angle INJ=90}$° and the line ${\displaystyle y=-c}$ is the ${\displaystyle directrix}$. If ${\displaystyle 4Kc<1}$, point ${\displaystyle L}$ is below the ${\displaystyle focus,\ \tan(\angle INJ)}$ is negative and ${\displaystyle 90}$° ${\displaystyle <\angle INJ<180}$°. ${\displaystyle \angle INJ}$ is obtuse and, as ${\displaystyle m}$ increases, ${\displaystyle \angle INJ}$ decreases, approaching ${\displaystyle 90}$°.

## Parabola and two tangents

Figure 5: Parabola and two tangents.
Origin ${\displaystyle (0,0)}$ at point ${\displaystyle O;}$ curve${\displaystyle :}$ ${\displaystyle y=Kx^{2};}$
point ${\displaystyle N}$${\displaystyle :\ (h,-c);}$ 2 tangents${\displaystyle :}$ ${\displaystyle NI,NJ;}$
chord ${\displaystyle IJ}$${\displaystyle :\ y=2Khx+c;}$ point ${\displaystyle L}$${\displaystyle :\ (0,c);}$

Refer to Figure 5.

The curve has equation: ${\displaystyle y=Kx^{2}.}$

Point ${\displaystyle N}$ with coordinates ${\displaystyle (h,-c)}$ is any point on the line ${\displaystyle y=-c.}$

Line ${\displaystyle NI}$ is tangent to the curve at point ${\displaystyle I\ (x_{1},y_{1})}$.

Line ${\displaystyle NJ}$ is tangent to the curve at point ${\displaystyle J\ (x_{2},y_{2})}$.

This section shows that the chord ${\displaystyle IJ}$ passes through the point ${\displaystyle (0,c).}$

Equation of any line through point ${\displaystyle N:\ y=mx-c-mh}$

Let this line intersect the curve:

${\displaystyle y=Kx^{2}=mx-c-mh}$

${\displaystyle Kx^{2}-mx+c+mh=0}$

${\displaystyle x={\frac {m\pm {\sqrt {m^{2}-4K(mh+c)}}}{2K}}={\frac {m\pm {\sqrt {m^{2}-4Kmh-4Kc}}}{2K}}}$

We want this line to be a tangent to the curve, therefore ${\displaystyle x}$ has exactly one value and the discriminant is ${\displaystyle 0}$:

{\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 ${\displaystyle R={\sqrt {K^{2}h^{2}+Kc}}}$

${\displaystyle m_{1}=2Kh-2R=}$ slope of tangent ${\displaystyle NI}$.

${\displaystyle m_{2}=2Kh+2R=}$ slope of tangent ${\displaystyle NJ}$.

{\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 ${\displaystyle IJ}$

{\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 ${\displaystyle IJ}$ on the ${\displaystyle Y}$ axis

{\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 ${\displaystyle INJ}$

If ${\displaystyle \angle INJ==90}$°

{\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 ${\displaystyle y={\frac {x^{2}}{4q}}}$ where the ${\displaystyle focus}$ has coordinates ${\displaystyle (0,q)}$.

${\displaystyle y=Kx^{2}\therefore K={\frac {1}{4q}}}$ or ${\displaystyle 4Kq=1.}$

If ${\displaystyle 4Kc==1,}$ then ${\displaystyle (0,c)}$ and ${\displaystyle (0,q)}$ are the same point, the chord ${\displaystyle IJ}$ passes through the ${\displaystyle focus}$ and the line ${\displaystyle y=-c}$ is the ${\displaystyle directrix}$.

## Area enclosed between parabola and chord

### Introduction

 Figure 6: The Parabola: ${\displaystyle y=x^{2}}$ Chord ${\displaystyle DC}$, parallel tangent ${\displaystyle AOB}$ and area ${\displaystyle DOCD.}$ ${\displaystyle \ \ \ \ \ \ base=CD=2.\ height=H_{1}T_{1}=1.}$ Chord ${\displaystyle OC}$, parallel tangent ${\displaystyle GHI}$ and area ${\displaystyle OHCO}$. ${\displaystyle \ \ \ \ \ \ base=OC={\sqrt {2}}.\ height=H_{2}T_{2}={\frac {\sqrt {2}}{8}}.}$ 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 ${\displaystyle AB,DC\ (H_{1}T_{1})}$ 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 ${\displaystyle OC=H_{2}T_{2}={\frac {1/2}{\sqrt {2}}}-{\frac {1/4}{\sqrt {2}}}={\frac {1}{4{\sqrt {2}}}}={\frac {\sqrt {2}}{8}}=height.}$ Length of ${\displaystyle OC={\sqrt {2}}=base.}$ Area between chord ${\displaystyle OC}$ and curve ${\displaystyle OHC={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\sqrt {2}})({\frac {\sqrt {2}}{8}})={\frac {2}{3}}({\frac {1}{4}})={\frac {1}{6}},}$ the same as that calculated earlier. These observations suggest that the area enclosed between chord and curve ${\displaystyle ={\frac {2}{3}}(base)(height).}$

### Proof

 Figure 7: The Parabola: ${\displaystyle y=Kx^{2}}$ Origin at point ${\displaystyle O:(0,0)}$ Points ${\displaystyle D,\ I:(p,0),(p,Kp^{2})}$ Points ${\displaystyle E,\ J:(q,0),(q,Kq^{2})}$ Chord ${\displaystyle IJ}$, parallel tangent ${\displaystyle FG}$ and area ${\displaystyle IOJI}$, the area enclosed between chord ${\displaystyle IJ}$ and curve. Length ${\displaystyle IJ=base.}$ Length ${\displaystyle HT=height.}$ 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 ${\displaystyle IJ.}$ ${\displaystyle y=K(p+q)x+c;\ \therefore c=y-K(p+q)x=Kqq-K(p+q)q=-Kpq}$ Equation of chord ${\displaystyle IJ:}$ ${\displaystyle y=K(p+q)x-Kpq.}$ 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 ${\displaystyle IOJ}$ ${\displaystyle ={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}-{\frac {Kqqq-Kppp}{3}}={\frac {Kqqq+3Kppq-3Kpqq-Kppp}{6}}}$ ${\displaystyle ={\frac {K(q-p)^{3}}{6}}={\frac {KS}{6}}.}$ The aim is to prove that: ${\displaystyle {\frac {2}{3}}(base)(height)={\frac {KS}{6}}}$ or ${\displaystyle {\frac {2}{3}}(L)({\frac {K(p-q)^{2}}{4R}})={\frac {KS}{6}}}$ or ${\displaystyle {\frac {2L(p-q)^{2}}{12R}}={\frac {S}{6}}}$ or ${\displaystyle L(p-q)^{2}=R(q-p)^{3}}$ or ${\displaystyle L=R(q-p)}$ where: ${\displaystyle L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}={\sqrt {(q-p)^{2}(K^{2}(q+p)^{2}+1)}}=(q-p){\sqrt {K^{2}(q+p)^{2}+1}}}$ ${\displaystyle R={\sqrt {K^{2}(p+q)^{2}+1}}}$ ${\displaystyle RHS=(q-p)R=(q-p){\sqrt {K^{2}(p+q)^{2}+1}}=L.}$ Therefore ${\displaystyle L=R(q-p)}$ 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.

### A worked example

Consider parabola: ${\displaystyle 16x^{2}-24xy+9y^{2}+20x-140y+600=0}$

and chord: ${\displaystyle 4x+3y-96=0.}$

The aim is to calculate area between chord and curve.

Calculate the points at which chord and parabola intersect: ${\displaystyle (5{\frac {7}{16}},24{\frac {3}{4}}),\ (15{\frac {1}{3}},11{\frac {5}{9}}).}$

#### Method 1. By chord and parallel tangent

 Figure 8: The Parabola: ${\displaystyle 16x^{2}-24xy+9y^{2}+20x-140y+600=0}$ Chord ${\displaystyle AD:4x+3y-96=0.}$ Length ${\displaystyle AD=base={\frac {2375}{144}}.}$ Parallel tangent ${\displaystyle CFB.\ height=BE={\frac {361}{24}}.}$ Area between chord and curve ${\displaystyle =DEAFD.}$ Length of chord {\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: ${\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y-19.2=0.}$ Equation of parallel tangent in normal form: ${\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y+g=0,}$ where ${\displaystyle g={\frac {4AFee-4BFde+4CFdd-DDee+2DEde-EEdd}{4AEe-2BDe-2BEd+4CDd}}}$ and ${\displaystyle d={\frac {4}{5}},\ e={\frac {3}{5}}}$ and ${\displaystyle A=16,\ B=-24,\ C=9,\ D=20,\ E=-140,\ F=600.}$ ${\displaystyle g=-{\frac {499}{120}}.}$ Equation of chord in normal form: ${\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y-19.2=0.}$ Equation of parallel tangent in normal form: ${\displaystyle {\frac {4}{5}}x+{\frac {3}{5}}y-{\frac {499}{120}}=0.}$ Distance between chord and parallel tangent ${\displaystyle =-{\frac {499}{120}}-(-19.2)={\frac {361}{24}}=height.}$ Area enclosed between chord and curve ${\displaystyle ={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\frac {2375}{144}})({\frac {361}{24}})={\frac {(2375)(361)}{3(12^{3})}}.}$

#### Method 2. By identifying the basic parabola.

 Figure 9: The Parabola: ${\displaystyle 16x^{2}-24xy+9y^{2}+20x-140y+600=0}$ Chord ${\displaystyle AD:4x+3y-96=0.}$ ${\displaystyle axis}$ is line ${\displaystyle FS_{1}S_{2}.\ directrix}$ is line ${\displaystyle RT.}$ ${\displaystyle focus}$ is point ${\displaystyle F:(2,6).}$ Line ${\displaystyle AS_{2}=p=-8{\frac {1}{2}}.}$ Line ${\displaystyle DS_{1}=q=7{\frac {1}{3}}}$ See Figure 9. Calculate directrix, focus and axis. Focus is distance ${\displaystyle 2}$ from directrix. Curve has the shape of ${\displaystyle y=Kx^{2}}$ where ${\displaystyle K={\frac {1}{4}}}$. Axis of symmetry has equation: ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+2=0.}$ Distance from point ${\displaystyle (5{\frac {7}{16}},24{\frac {3}{4}})}$ to axis of symmetry ${\displaystyle =-8{\frac {1}{2}}=p.}$ Distance from point ${\displaystyle (15{\frac {1}{3}},11{\frac {5}{9}})}$ to axis of symmetry ${\displaystyle =7{\frac {1}{3}}=q.}$ ${\displaystyle q-p=7{\frac {1}{3}}-(-8{\frac {1}{2}})=15{\frac {5}{6}}={\frac {95}{6}}.}$ Area between chord and curve {\displaystyle {\begin{aligned}=&\ {\frac {K(q-p)^{3}}{6}}\\\\=&\ {\frac {1}{24}}({\frac {95}{6}})({\frac {95}{6}})({\frac {95}{6}})\\\\=&\ {\frac {(2375)(361)}{3(12^{3})}}.\end{aligned}}} or: ${\displaystyle height={\frac {K(p-q)^{2}}{4R}}}$ where ${\displaystyle R={\sqrt {K^{2}(p+q)^{2}+1}}}$ ${\displaystyle (p-q)^{2}={\frac {95(95)}{36}}}$ ${\displaystyle (p+q)^{2}={\frac {49}{36}}}$ ${\displaystyle R={\sqrt {{\frac {49}{16(36)}}+1}}={\sqrt {{\frac {7^{2}}{24^{2}}}+{\frac {24^{2}}{24^{2}}}}}={\sqrt {\frac {25^{2}}{24^{2}}}}={\frac {25}{24}}}$ ${\displaystyle height={\frac {\frac {95(95)}{(4)36}}{4({\frac {25}{24}})}}={\frac {\frac {95(95)}{(4)36}}{\frac {25}{6}}}=({\frac {95(95)}{(4)36}})({\frac {6}{25}})={\frac {19(19)}{24}}={\frac {361}{24}}.}$ Calculate ${\displaystyle base}$ as above and area enclosed between chord and curve ${\displaystyle ={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\frac {2375}{144}})({\frac {361}{24}})={\frac {(2375)(361)}{3(12^{3})}}}$.