Parabola

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

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

See Figure 1.

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

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

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

Let an arbitrary point on the curve be $(x,y)$.

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

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

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

If the directrix is parallel to the $X$ 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 $ax+by+c=0$ where at least one of $a,b$ is non-zero.

Let the focus be $(p,q)$.

Let $(x,y)$ be any point on the curve.

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

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

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

By definition these two lengths are equal:

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

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

Square both sides:

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

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

Expand and the result is:

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

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

${\begin{aligned}A&=b^{2}\\B&=-2ab\\C&=a^{2}\\D&=-2(ac+pR)\\E&=-2(bc+qR)\\F&=R(p^{2}+q^{2})-c^{2}\\\\B&^{2}-4AC=0\\R&=a^{2}+b^{2}\end{aligned}}$

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

An Example

Figure 2: The Parabola $16x^{2}+24xy+9y^{2}-20x-140y+100=0$ Green line is $directrix:\ 3x-4y-5=0$ Blue line is $axis:\ 4x+3y-10=0$ Focus at point $F:\ (1,2)$ Vertex at point $V:\ (1.6,1.2)$ $VF=1.$ Shape of curve is: $y={\frac {x^{2}}{4}}.$ See Figure 2. $(a,b,c)=(3,-4,-5)$ $(p,q)=(1,2)$ The equation of the parabola is derived as follows: ${\begin{aligned}&A=b^{2}=(-4)^{2}=16\\&B=-2ab=-2(3)(-4)=24\\&C=a^{2}=(3)^{2}=9\\&R=a^{2}+b^{2}=25\\&D=-2(ac+pR)=-2((3)(-5)+(1)(25))=-2(-15+25)=-2(10)=-20\\&E=-2(bc+qR)=-2((-4)(-5)+(2)(25))=-2(20+50)=-2(70)=-140\\&F=R(p^{2}+q^{2})-c^{2}=25(1+4)-25=100\end{aligned}}$ The equation of the parabola in Figure 2 is: $16x^{2}+24xy+9y^{2}-20x-140y+100=0.$ Equation of directrix in normal form: ${\frac {3}{5}}x-{\frac {4}{5}}y-1=0.$ Distance from $focus$ to $directrix={\frac {3}{5}}(1)-{\frac {4}{5}}(2)-1=-2.$ Distance from vertex to focus $=1={\frac {1}{4K}}$. Therefore, curve has shape of $y=Kx^{2}$ where $K={\frac {1}{4}}$. Caution: An interesting situation occurs if the focus is on the directrix. Consider the directrix: $4x-3y+15=0$ and the focus $(3,9)$ which is on the directrix. $a=4,\ b=-3,\ c=15,\ p=3,\ q=9$ In this case the "parabola" has equation: $9xx+24xy+16yy-270x-360y+2025=0$. This seems to be the equation of a parabola because $B^{2}-4AC=0$, but look closely. $9xx+24xy+16yy-270x-360y+2025=(3x+4y-45)^{2}$. The result is a line through the focus and normal to the directrix. If you solve for $p,q,c$ using the algebraic solutions, you will produce the values $3,9,15$ as above. However, the distance between focus and directrix = ${\frac {4}{5}}x+{\frac {-3}{5}}y+3$ where $x=p;\ y=q;$ distance $={\frac {4}{5}}(3)+{\frac {-3}{5}}(9)+3=0.$

Reverse-Engineering the Parabola

Figure 3: Parabola with 2 tangents parallel to axes. Tangent $ABC:y={\frac {16}{3}}$. Tangent $ADE:x=-0.25$. Point $A$ on directrix, oblique, thin, black line.${\displaystyle }$ Line $AFG$ perpendicular to focal chord $BFD.$ Focus at $F:(1,7)$. Given a parabola in form $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$ the aim is to produce the directrix and the focus. We will solve the example shown in Figure 3: $9x^{2}-24xy+16y^{2}+70x-260y+1025=0$, where: ${\begin{aligned}A&=9\\B&=-24\\C&=16\\D&=70\\E&=-260\\F&=1025\end{aligned}}$ $a={\sqrt {C}}=4;\ b={\frac {-B}{2a}}={\frac {-(-24)}{8}}=3$. 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 $x$. $Ax^{2}+Bxy+Dx+Cy^{2}+Ey+F=0$ $Ax^{2}+(By+D)x+(Cy^{2}+Ey+F)=0$ At the tangent there is exactly one value of $x$. Therefore the discriminant must be $0$. $(By+D)^{2}-4(A)(Cy^{2}+Ey+F)=0$ $BByy+2BDy+DD-4ACyy-4AEy-4AF=0$ $(BB-4AC)yy+(2BD-4AE)y+(DD-4AF)=0$ In the general parabola $B^{2}-4AC=0$ therefore $y={\frac {4AF-DD}{2BD-4AE}}$. In this example $y={\frac {16}{3}};\ x={\frac {-(By+D)}{2A}}={\frac {29}{9}}$. Point $B(x_{1},y_{1})$ has coordinates $({\frac {29}{9}},\ {\frac {16}{3}})$. The line $ABC$ is tangent to the curve at $B$ and has equation: $y={\frac {16}{3}}$. Put the equation of the parabola in the form of a quadratic in $y$: $Cy^{2}+(Bx+E)y+(Ax^{2}+Dx+F)=0$. By using calculations similar to the above, $x={\frac {4CF-EE}{2BE-4CD}}=-0.25$ and $y={\frac {-(Bx+E)}{2C}}=7.9375$. Point $D(x_{2},y_{2})$ has coordinates $(-0.25,\ 7.9375)$. The line $ADE$ is tangent to the curve at $D$ and has equation: $x=-0.25$. Point $A$ at the intersection of the two tangents has coordinates $(x_{2},y_{1})=(-0.25,\ {\frac {16}{3}})$, and point $A$ is on the directrix, the equation of which is: $ax+by+c=0$. Put known values into the equation of the directrix: $(4)(-0.25)+(3){\frac {16}{3}}+c=0$. Therefore $c=-15$ and the equation of the directrix is: $4x+3y-15=0$. The coordinates of points $B,D$ are known. Therefore chord $BD$ is defined as: ${\frac {3}{5}}x+{\frac {4}{5}}y-6.2=0$. Draw the line $AG$ perpendicular to $BD$. The line $AG$ is defined as: ${\frac {4}{5}}x-{\frac {3}{5}}y+3.4=0$. Point $F$ at the intersection of lines $BD,AG$ is the focus with coordinates $(1,7)$. Method 2. By algebra ${\begin{aligned}A&=b^{2}\\B&=-2ab\\C&=a^{2}\\D&=-2(ac+pR)\\E&=-2(bc+qR)\\F&=R(p^{2}+q^{2})-c^{2}\\\\B&^{2}-4AC=0\\R&=a^{2}+b^{2}\end{aligned}}$ After rearranging the above values, there are three equations to be solved for three unknowns: $p,q,c$: ${\begin{aligned}D+2ac+2pR=0\\E+2bc+2qR=0\\F-Rpp-Rqq+cc=0\end{aligned}}$ The solutions are: $c={\frac {4FR-(DD+EE)}{4Da+4Eb}}$ $p={\frac {-(D+2ac)}{2R}}$ $q={\frac {-(E+2bc)}{2R}}$ If $b$ is $0,\ A=B=0,\ R=a^{2}=C,$ the parabola becomes the quadratic: $-Dx=Cy^{2}+Ey+F$ and: ${\begin{aligned}p&={\frac {E^{2}-D^{2}-4FC}{4CD}}\\q&={\frac {-E}{2C}}\\c&={\frac {4FC-(DD+EE)}{4Da}}\end{aligned}}$ The directrix has equation: $ax+c=0;\ ax=-c;\ x={\frac {DD+EE-4FC}{4CD}}$. If $D$ is $-1$, then: ${\begin{aligned}x&=Cy^{2}+Ey+F\\p&={\frac {E^{2}-1-4FC}{4C(-1)}}={\frac {1-(E^{2}-4CF)}{4C}}\\q&={\frac {-E}{2C}}\end{aligned}}$ The directrix has equation: $x={\frac {-1-(E^{2}-4CF)}{4C}}$. If $a$ is $0,\ B=C=0,\ R=b^{2}=A,$ the parabola becomes the quadratic: $-Ey=Ax^{2}+Dx+F$ and: ${\begin{aligned}p&={\frac {-D}{2A}}\\q&={\frac {D^{2}-E^{2}-4FA}{4EA}}\\c&={\frac {4FA-(DD+EE)}{4Eb}}\end{aligned}}$ The directrix has equation: $by+c=0;\ by=-c;\ y={\frac {DD+EE-4FA}{4EA}}$. 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. If $E$ is $-1$, then: ${\begin{aligned}y&=Ax^{2}+Dx+F\\p&={\frac {-D}{2A}}\\q&={\frac {D^{2}-1-4FA}{4(-1)A}}={\frac {1-(D^{2}-4AF)}{4A}}\end{aligned}}$ The directrix has equation: $y={\frac {-1-(D^{2}-4AF)}{4A}}$. These values agree with the corresponding values in the graph of $y=x^{2}-2x-3$ to the right.

Slope of the Parabola

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

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

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

${\begin{aligned}(+A+Bm+Cmm)&xx+\\(-BXm+BY-2CXmm+2CYm+D+Em)&x+\\(+CXXmm-2CXYm+CYY-EXm+EY+F)&=0\end{aligned}}$

We want the line to be tangent to the curve. Therefore $x$ must have exactly one value and the discriminant is $0.$

Discriminant =

${\begin{aligned}(-BXm+BY-2CXmm+2CYm+D+Em)(-BXm+BY-2CXmm+2CYm+D+Em)&\\-4(+A+Bm+Cmm)(+CXXmm-2CXYm+CYY-EXm+EY+F)&=0\end{aligned}}$

The above discriminant is a quadratic in $m$:

${\begin{aligned}(-4ACXX+BBXX+2BEX-4CDX-4CF+EE)&mm+\\(+8ACXY+4AEX-2BBXY-2BDX-2BEY-4BF+4CDY+2DE)&m+\\(-4ACYY-4AEY-4AF+BBYY+2BDY+DD)&=0\end{aligned}}$

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

where:

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

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

${\begin{aligned}(AX&X+BXY+CYY+DX+EY+F)=0;\ R=0\\\\m=&\ {\frac {-(+8ACXY+4AEX-2BBXY-2BDX-2BEY-4BF+4CDY+2DE)}{2(-4ACXX+BBXX+2BEX-4CDX-4CF+EE)}}\\\\=&\ {\frac {2BBXY-8ACXY+2BDX-4AEX+2BEY-4CDY+4BF-2DE}{2(BBXX-4ACXX+2BEX-4CDX+EE-4CF)}}\\\\=&\ {\frac {(BB-4AC)XY+(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(BB-4AC)XX+(2BE-4CD)X+EE-4CF}}\\\\=&\ {\frac {(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(2BE-4CD)X+EE-4CF}}\end{aligned}}$

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

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

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

Let the equation of a line be: $x=My+c$ in which case $M={\frac {1}{m}}.$ By using calculations similar to the above it can be shown that: $M={\frac {(BD-2AE)X+(BE-2CD)Y+2BF-DE}{(2BD-4AE)Y+DD-4AF}}$. $m={\frac {1}{M}}$ therefore: $m={\frac {(2BD-4AE)Y+DD-4AF}{(BD-2AE)X+(BE-2CD)Y+2BF-DE}}$ When slope is displayed in this format, we see that slope is zero if $(2BD-4AE)Y+DD-4AF=0$. The line $y={\frac {4AF-DD}{2BD-4AE}}$ is tangent to the curve.

This examination of the parabola has produced two expressions for slope of the parabola: $m={\frac {(2BD-4AE)y+DD-4AF}{(BD-2AE)x+(BE-2CD)y+2BF-DE}}={\frac {(BD-2AE)x+(BE-2CD)y+2BF-DE}{(2BE-4CD)x+EE-4CF}}$. where the point $(x,y)$ is any arbitrary point on the curve. Therefore $m=\pm {\sqrt {\frac {(2BD-4AE)y+DD-4AF}{(2BE-4CD)x+EE-4CF}}}$. This formula for $m$ contains both tangents parallel to the axes and is derived without calculus.

The formula from calculus below is simpler and unambiguous concerning sign. ${\begin{aligned}Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\\\\{\frac {d}{dx}}(Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F)=0\\\\A(2x)+B(x{\frac {dy}{dx}}+y{\frac {dx}{dx}})+C(2y{\frac {dy}{dx}})+D+E({\frac {dy}{dx}})=0\\\\A(2x)+B(x{\frac {dy}{dx}}+y)+C(2y{\frac {dy}{dx}})+D+E({\frac {dy}{dx}})=0\\\\2Ax+Bx{\frac {dy}{dx}}+By+2Cy{\frac {dy}{dx}}+D+E{\frac {dy}{dx}}=0\\\\{\frac {dy}{dx}}(Bx+2Cy+E)=-(2Ax+By+D)\\\\m={\frac {dy}{dx}}={\frac {-(2Ax+By+D)}{Bx+2Cy+E}}\end{aligned}}$

The slope of the parabola is $0$ where $(2BD-4AE)y+DD-4AF=0;\ 2Ax+By+D=0;$ or: $y={\frac {4AF-DD}{2BD-4AE}};\ x={\frac {-(By+D)}{2A}}.$ The slope of the parabola is vertical where $(2BE-4CD)x+EE-4CF=0;\ Bx+2Cy+E=0;$ or: $x={\frac {4CF-EE}{2BE-4CD}};\ y={\frac {-(Bx+E)}{2C}}.$

Caution: If the curve is $y=Kx^{2}$, the slope can never be vertical. If the curve is $x=Ky^{2}$, the slope can never be $0$.

If $B=C=0$ and $E=-1$, the equation of the parabola becomes: $y=Ax^{2}+Dx+F$ and: ${\begin{aligned}m=&\ {\frac {(2BD-4AE)y+DD-4AF}{(BD-2AE)x+(BE-2CD)y+2BF-DE}}\\\\=&\ {\frac {(-4A(-1))y+DD-4AF}{(-2A(-1))x-D(-1)}}={\frac {4Ay+DD-4AF}{2Ax+D}}\\\\=&\ {\frac {4A(Ax^{2}+Dx+F)+DD-4AF}{2Ax+D}}={\frac {4A^{2}x^{2}+4ADx+4AF+DD-4AF}{2Ax+D}}\\\\=&\ {\frac {4A^{2}x^{2}+4ADx+DD}{2Ax+D}}\\\\=&\ 2Ax+D\\\\m=&\ {\frac {(BD-2AE)x+(BE-2CD)y+2BF-DE}{(2BE-4CD)x+EE-4CF}}\\\\=&\ {\frac {((0)D-2A(-1))x+((0)E-2(0)D)y+2(0)F-D(-1)}{(2(0)E-4(0)D)x+(-1)(-1)-4(0)F}}\\\\=&\ {\frac {(-2A(-1))x-D(-1)}{(-1)(-1)}}\\\\=&\ 2Ax+D\\\\m=&\ {\frac {-(2Ax+By+D)}{Bx+2Cy+E}}\\\\=&\ {\frac {-(2Ax+(0)y+D)}{(0)x+2(0)y+(-1)}}\\\\=&\ 2Ax+D\end{aligned}}$

Point at given slope

Given parabola defined by $A,B,C,D,E,F$ and slope $m={\frac {yvalue}{xvalue}}$ where at least one of $yvalue,xvalue$ is non-zero, calculate point at which the slope is $m.$ Let $m={\frac {-(2Ax+By+D)}{Bx+2Cy+E}}={\frac {s}{t}}={\frac {yvalue}{xvalue}}$ Then $s(Bx+2Cy+E)+t(2Ax+By+D)=0.$ Let $G=(Bs+2At);\ H=(2Cs+Bt);\ I=(Es+Dt).$ Then $Gx+Hy+I=0\ \dots \ (1)$ Substitute in the equation of the parabola and $y={\frac {-(AII-DGI+FGG)}{(2AHI-BGI-DGH+EGG)}}\ \dots \ (2)$ As shown below, with a little manipulation of the data, the same formula can be used to calculate $x.$ Equation $(1)$ above is the equation of a straight line with slope ${\frac {-G}{H}}.$ Substitute for $G,H$ and the slope of $(1)$ becomes ${\frac {b}{a}}.$ Equation $(1)$ is that of a line parallel to the axis of symmetry of the parabola. It is possible for both both $G,H$ to equal $0$ in which case the calculation of $y,(2)$ above fails as an attempt to divide by $0.$ 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 $9x^{2}-24xy+16y^{2}+70x-260y+1025=0.$ 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 $(0,0)$ at point $O;$ curve$:$ $y=Kx^{2};$
chord $IJ$$:\ y=mx+c;$ point $L$$:\ (0,c);$
2 tangents$:$ $IN,JN;$ point $N$$:\ ({\frac {m}{2K}},-c)$

Refer to Figure 4.

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

The chord $IJ$ has equation: $y=mx+c$.

Point $L$ has coordinates $(0,c).$

Line $IN$ is tangent to the curve at $I.$

Line $JN$ is tangent to the curve at $J.$

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

${\begin{aligned}y=&\ Kx^{2}=mx+c\\\\Kx^{2}&-mx-c=0\\\\x=&\ {\frac {m\pm {\sqrt {m^{2}+4Kc}}}{2K}}\\\\=&\ {\frac {m\pm R}{2K}}\end{aligned}}$

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

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

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

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

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

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

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

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

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

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

Equation of tangent $IN:$

${\begin{aligned}y=&\ s_{1}x+c_{1}\\c_{1}=&\ y_{1}-s_{1}x_{1}={\frac {m^{2}-mR+2Kc}{2K}}-(m-R)({\frac {m-R}{2K}})={\frac {-(m^{2}-mR+2Kc)}{2K}}\\y=&\ (m-R)x-{\frac {m^{2}-mR+2Kc}{2K}}\end{aligned}}$

Equation of tangent $JN:$

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

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

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

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

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

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

The $Y$ coordinate of point $N:$

${\displaystyle }$

${\begin{aligned}y=&\ s_{1}({\frac {m}{2K}})+c_{1}\\=&\ (m-R)({\frac {m}{2K}})-{\frac {m^{2}-mR+2Kc}{2K}}\\=&\ {\frac {m^{2}-mR}{2K}}-{\frac {m^{2}-mR+2Kc}{2K}}\\=&\ {\frac {-2Kc}{2K}}\\=&\ -c\end{aligned}}$

Any chord that passes through the point $L\ (0,c)$ has two tangents that intersect on the line $y=-c$.

Angle $INJ$

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

Parabola and two tangents

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

Refer to Figure 5.

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

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

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

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

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

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

Let this line intersect the curve:

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

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

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

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

${\begin{aligned}m^{2}-&4Kmh-4Kc=0\\\\m=&\ {\frac {4Kh\pm {\sqrt {(4Kh)^{2}-4(1)(-4Kc)}}}{2}}\\\\=&\ {\frac {4Kh\pm {\sqrt {16K^{2}h^{2}+16Kc}}}{2}}\\\\=&\ 2Kh\pm 2R\end{aligned}}$

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

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

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

${\begin{aligned}x=&\ {\frac {m}{2K}}\\\\x_{1}=&\ {\frac {m_{1}}{2K}}={\frac {2Kh-2R}{2K}}={\frac {Kh-R}{K}}\\\\y_{1}=&\ Kx_{1}^{2}=K({\frac {Kh-R}{K}})({\frac {Kh-R}{K}})={\frac {(Kh-R)(Kh-R)}{K}}\\\\x_{2}=&\ {\frac {Kh+R}{K}}\\\\y_{2}=&\ {\frac {(Kh+R)(Kh+R)}{K}}\end{aligned}}$

Slope of chord $IJ$

${\begin{aligned}=&\ {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\\\\=&\ ({\frac {(Kh+R)(Kh+R)}{K}}-{\frac {(Kh-R)(Kh-R)}{K}})/({\frac {Kh+R}{K}}-{\frac {Kh-R}{K}})\\\\=&\ ({\frac {KKhh+2KhR+RR-(KKhh-2KhR+RR)}{K}})/({\frac {2R}{K}})\\\\=&\ ({\frac {4KhR}{K}})({\frac {K}{2R}})\\\\=&\ 2Kh\end{aligned}}$

Intercept of chord $IJ$ on the $Y$ axis

${\begin{aligned}=&\ y_{1}-2Khx_{1}\\\\=&\ {\frac {(Kh-R)(Kh-R)}{K}}-2Kh{\frac {Kh-R}{K}}\\\\=&\ {\frac {KKhh-2KhR+RR}{K}}-{\frac {2Kh(Kh-R)}{K}}\\\\=&\ {\frac {KKhh-2KhR+RR-2KhKh+2KhR}{K}}\\\\=&\ {\frac {KKhh+RR-2KhKh}{K}}\\\\=&\ {\frac {-KKhh+KKhh+Kc}{K}}\\\\=&\ {\frac {+Kc}{K}}\\\\=&\ c\end{aligned}}$

Angle $INJ$

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If $\angle INJ==90$°

${\begin{aligned}&(m_{1})(m_{2})=-1\\\\&(2Kh-2R)(2Kh+2R)=-1\\\\&4(KKhh-RR)=-1\\\\&4(KKhh-(KKhh+Kc))=-1\\\\&4(-Kc)=-1\\\\&4Kc=1\end{aligned}}$

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

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

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

Area enclosed between parabola and chord

Introduction

Figure 6: The Parabola: $y=x^{2}$ Chord $DC$, parallel tangent $AOB$ and area $DOCD.$ $\ \ \ \ \ \ base=CD=2.\ height=H_{1}T_{1}=1.$ Chord $OC$, parallel tangent $GHI$ and area $OHCO$. $\ \ \ \ \ \ base=OC={\sqrt {2}}.\ height=H_{2}T_{2}={\frac {\sqrt {2}}{8}}.$ See Figure 6. The curve is: $y=x^{2}$. Integral is: ${\frac {x^{3}}{3}}$. Area under curve $(OBC)$ ${\begin{aligned}x=&1\\=\ \ \ \ \ &[{\frac {x^{3}}{3}}]={\frac {1}{3}}\\x=&0\end{aligned}}$ Area under curve $(DOC)=$ area$(OAD)+$ area$(OBC)={\frac {2}{3}}$ Area between chord $DC$ and curve $DOC=2-{\frac {2}{3}}={\frac {4}{3}}$. Consider the chord $CD$. Call this the $base$ with value $2$. The tangent $AOB$ through the origin is parallel to $base\ (DC)$, and the perpendicular distance between $AB,DC\ (H_{1}T_{1})$ is $1$. Call this distance the $height$ with value $1$. In this case the area enclosed between chord $DC$ and curve $DOC={\frac {2}{3}}(base)(height)={\frac {2}{3}}(2)(1)={\frac {4}{3}},$ the same as that calculated earlier. Consider the chord $OC$ and curve $OHC.$ By inspection the area between chord $OC$ and curve $OHC={\frac {1}{2}}-{\frac {1}{3}}={\frac {1}{6}}.$ Chord $OC$ has equation $y=x;\ x-y=0;\ {\frac {x}{\sqrt {2}}}-{\frac {y}{\sqrt {2}}}=0$ in normal form. The line $GHI$ is parallel to $base\ OC$ and touches the curve at $H({\frac {1}{2}},\ {\frac {1}{4}}).$ Distance from $H$ to chord $OC=H_{2}T_{2}={\frac {1/2}{\sqrt {2}}}-{\frac {1/4}{\sqrt {2}}}={\frac {1}{4{\sqrt {2}}}}={\frac {\sqrt {2}}{8}}=height.$ Length of $OC={\sqrt {2}}=base.$ Area between chord $OC$ and curve $OHC={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\sqrt {2}})({\frac {\sqrt {2}}{8}})={\frac {2}{3}}({\frac {1}{4}})={\frac {1}{6}},$ the same as that calculated earlier. These observations suggest that the area enclosed between chord and curve $={\frac {2}{3}}(base)(height).$

Proof

Figure 7: The Parabola: $y=Kx^{2}$ Origin at point $O:(0,0)$ Points $D,\ I:(p,0),(p,Kp^{2})$ Points $E,\ J:(q,0),(q,Kq^{2})$ Chord $IJ$, parallel tangent $FG$ and area $IOJI$, the area enclosed between chord $IJ$ and curve. Length $IJ=base.$ Length $HT=height.$ We prove this identity for the general case. See Figure 7. Slope of chord $IJ={\frac {Kqq-Kpp}{q-p}}={\frac {K((q+p)(q-p))}{q-p}}=K(q+p).$ Find equation of chord $IJ.$ $y=K(p+q)x+c;\ \therefore c=y-K(p+q)x=Kqq-K(p+q)q=-Kpq$ Equation of chord $IJ:$ $y=K(p+q)x-Kpq.$ Find equation of tangent $FG.$ ${\begin{aligned}&y=Kx^{2}\\&y=K(p+q)x+c\\\therefore \ &Kx^{2}=K(p+q)x+c\\&Kx^{2}-K(p+q)x-c=0\end{aligned}}$ We choose a value of $c$ that gives $x$ exactly one value. Therefore discriminant $K^{2}(p+q)^{2}+4Kc=0;\ c={\frac {-K(p+q)^{2}}{4}}.$ $y=K(p+q)x-{\frac {K(p+q)^{2}}{4}};\ K(p+q)x-y-{\frac {K(p+q)^{2}}{4}}=0;$ Equation of tangent $FG$ in normal form: ${\frac {K(p+q)x-y-{\frac {K(p+q)^{2}}{4}}}{\sqrt {K^{2}(p+q)^{2}+1}}}=0.$ Equation of chord $IJ$ in normal form: ${\frac {K(p+q)x-y-Kpq}{\sqrt {K^{2}(p+q)^{2}+1}}}=0.$ Therefore distance between chord $IJ$ and tangent $FG$ $=height={\frac {-Kpq-(-{\frac {K(p+q)^{2}}{4}})}{R}}={\frac {K(p+q)^{2}-4Kpq}{4R}}={\frac {K(p-q)^{2}}{4R}}$ where $R={\sqrt {K^{2}(p+q)^{2}+1}}.$ Length of chord $IJ=base=L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}.$ Area under chord $IJ=(q-p){\frac {Kqq+Kpp}{2}}={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}$ Area under curve $IOJ$ ${\begin{aligned}x=&q\\=\ \ \ \ \ &[{\frac {Kx^{3}}{3}}]={\frac {Kqqq-Kppp}{3}}\\x=&p\end{aligned}}$ Area between chord $IJ$ and curve $IOJ$ $={\frac {Kqqq+Kppq-Kqqp-Kppp}{2}}-{\frac {Kqqq-Kppp}{3}}={\frac {Kqqq+3Kppq-3Kpqq-Kppp}{6}}$ $={\frac {K(q-p)^{3}}{6}}={\frac {KS}{6}}.$ The aim is to prove that: ${\frac {2}{3}}(base)(height)={\frac {KS}{6}}$ or ${\frac {2}{3}}(L)({\frac {K(p-q)^{2}}{4R}})={\frac {KS}{6}}$ or ${\frac {2L(p-q)^{2}}{12R}}={\frac {S}{6}}$ or $L(p-q)^{2}=R(q-p)^{3}$ or $L=R(q-p)$ where: $L={\sqrt {(Kqq-Kpp)^{2}+(q-p)^{2}}}={\sqrt {(q-p)^{2}(K^{2}(q+p)^{2}+1)}}=(q-p){\sqrt {K^{2}(q+p)^{2}+1}}$ $R={\sqrt {K^{2}(p+q)^{2}+1}}$ $RHS=(q-p)R=(q-p){\sqrt {K^{2}(p+q)^{2}+1}}=L.$ Therefore $L=R(q-p)$ and area enclosed between curve and chord =${\frac {2}{3}}(base)(height)$ where $base$ is the length of the chord, and $height$ is the perpendicular distance between chord and tangent parallel to chord.

A worked example

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

and chord: $4x+3y-96=0.$

The aim is to calculate area between chord and curve.

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

Method 1. By chord and parallel tangent

Figure 8: The Parabola: $16x^{2}-24xy+9y^{2}+20x-140y+600=0$ Chord $AD:4x+3y-96=0.$ Length $AD=base={\frac {2375}{144}}.$ Parallel tangent $CFB.\ height=BE={\frac {361}{24}}.$ Area between chord and curve $=DEAFD.$ Length of chord ${\begin{aligned}=&\ {\sqrt {(15{\frac {1}{3}}-5{\frac {7}{16}})^{2}+(11{\frac {5}{9}}-24{\frac {3}{4}})^{2}}}\\\\=&\ {\sqrt {({\frac {46}{3}}-{\frac {87}{16}})^{2}+({\frac {99}{4}}-{\frac {104}{9}})^{2}}}\\\\=&\ {\sqrt {({\frac {475}{48}})^{2}+({\frac {475}{36}})^{2}}}=\ {\sqrt {({\frac {475(3)}{48(3)}})^{2}+({\frac {475(4)}{36(4)}})^{2}}}\\\\=&\ {\sqrt {({\frac {475(3)}{144}})^{2}+({\frac {475(4)}{144}})^{2}}}=\ {\frac {475(5)}{144}}\\\\=&\ {\frac {2375}{144}}=base.\end{aligned}}$ Equation of chord in normal form: ${\frac {4}{5}}x+{\frac {3}{5}}y-19.2=0.$ Equation of parallel tangent in normal form: ${\frac {4}{5}}x+{\frac {3}{5}}y+g=0,$ where $g={\frac {4AFee-4BFde+4CFdd-DDee+2DEde-EEdd}{4AEe-2BDe-2BEd+4CDd}}$ and $d={\frac {4}{5}},\ e={\frac {3}{5}}$ and $A=16,\ B=-24,\ C=9,\ D=20,\ E=-140,\ F=600.$ $g=-{\frac {499}{120}}.$ Equation of chord in normal form: ${\frac {4}{5}}x+{\frac {3}{5}}y-19.2=0.$ Equation of parallel tangent in normal form: ${\frac {4}{5}}x+{\frac {3}{5}}y-{\frac {499}{120}}=0.$ Distance between chord and parallel tangent $=-{\frac {499}{120}}-(-19.2)={\frac {361}{24}}=height.$ Area enclosed between chord and curve $={\frac {2}{3}}(base)(height)={\frac {2}{3}}({\frac {2375}{144}})({\frac {361}{24}})={\frac {(2375)(361)}{3(12^{3})}}.$

Method 2. By identifying the basic parabola.

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