Parabola



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 $$D_1D_4$$$$:\ y = -q.$$ 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 $$P_2FP_4$$ is a chord. Because chord $$P_2FP_4$$ 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:

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

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

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

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

An Example


See Figure 2.

$$(a,b,c) = (3,-4,-5)$$

$$(p,q) = (1,2)$$

The equation of the parabola is derived as follows:

$$\begin{align} &A = b^2 = (-4)^2 = 16\\ &B = -2ab = -2(3)(-4) = 24\\ &C = a^2 = (3)^2 = 9\\ &R = a^2 + b^2 = 25\\ &D = -2(ac + pR) = -2((3)(-5) + (1)(25)) = -2(-15+25) = -2(10) = -20\\ &E = -2(bc + qR) = -2((-4)(-5) + (2)(25)) = -2(20 + 50) = -2(70) = -140\\ &F = R(p^2 + q^2) - c^2 = 25(1 + 4) - 25 = 100 \end{align}$$

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


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:

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

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

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{align}

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{align}$$

After rearranging the above values, there are three equations to be solved for three unknowns: $$p, q, c$$:


 * $$\begin{align}

D + 2ac + 2pR = 0\\ E + 2bc + 2qR = 0\\ F - Rpp - Rqq + cc = 0

\end{align}$$

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{align}

p& = \frac{E^2 - D^2 -4FC}{4CD}\\ q& = \frac{-E}{2C}\\ c& = \frac{  4FC - (DD + EE )  }{  4Da   } \end{align}$$

The directrix has equation: $$ax + c = 0;\ ax = -c;\ x = \frac{DD+EE - 4FC}{4CD}$$.

If $$D$$ is $$-1$$, then:


 * $$\begin{align}

x& = Cy^2 + Ey + F\\ p& = \frac{E^2 - 1 -4FC}{4C(-1)} = \frac{1 - (E^2 - 4CF)}{4C}\\ q& = \frac{-E}{2C} \end{align}$$

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{align}

p& = \frac{-D}{2A}\\ q& = \frac{D^2 - E^2 - 4FA}{4EA}\\ c& = \frac{  4FA - (DD + EE )  }{  4Eb    } \end{align}$$

The directrix has equation: $$by + c = 0;\ by = -c;\ y = \frac{ DD + EE - 4FA }{ 4EA  }$$.

[[File:Quadratic function graph key values.svg|thumb|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{align}

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{align}$$

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{align}

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

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

Discriminant =


 * $$\begin{align}

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

The above discriminant is a quadratic in $$m$$:


 * $$\begin{align}

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

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

where:


 * $$\begin{align}

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

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


 * $$\begin{align}

(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{align}$$

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{align}

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{align}$$

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{align}

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^2x^2+4ADx+4AF + DD - 4AF } {  2Ax + D }\\ \\ =&\ \frac { 4A^2x^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{align}$$

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.

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.

Calculate point at which tangent is vertical.

Parabola and any chord


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{align}

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{align}$$

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{align}

y =&\ s_1x+c_1\\ c_1 =&\ y_1-s_1x_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{align}$$

Equation of tangent $$JN:$$


 * $$\begin{align}

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

\end{align}$$

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 $$N:$$

$$$$


 * $$\begin{align}

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{align}$$

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

Angle $$INJ$$

Using:


 * $$\begin{align}

\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{align}$$

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


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{align}

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

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

$$m_1=2Kh-2R=$$ slope of tangent $$NI$$.

$$m_2=2Kh+2R =$$ slope of tangent $$NJ$$.


 * $$\begin{align}

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{align}$$

Slope of chord $$IJ$$


 * $$\begin{align}

=&\ \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{align}$$

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


 * $$\begin{align}

=&\ 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{align}$$

Angle $$INJ$$ --

If $$\angle INJ == 90$$°


 * $$\begin{align}

&(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{align}$$

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 $$directrix$$.

Introduction


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

Area under curve $$(OBC)$$


 * $$\begin{align}

x =& 1\\ =\ \ \ \ \ &[\frac{x^3}{3}] = \frac{1}{3}\\ x =& 0 \end{align}$$

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 $$AB, DC\ (H_1T_1)$$ 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 $$OC = H_2T_2 = \frac{1/2}{\sqrt{2}} - \frac{1/4}{\sqrt{2}} = \frac{1}{4\sqrt{2}} = \frac{\sqrt{2}}{8} = height.$$

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


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{align}

&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{align}$$

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{align}

x =& q\\ =\ \ \ \ \ &[\frac{Kx^3}{3}] = \frac{Kqqq-Kppp}{3}\\ x =& p \end{align}$$

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


Length of chord


 * $$\begin{align}

=&\ \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{align}$$

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.
[[File:20170526 Determine basic parabola 00.png|thumb|300px| Figure 9: The Parabola: $$16x^2 - 24xy + 9y^2 + 20x - 140y + 600 = 0$$ Chord $$AD: 4x + 3y - 96 = 0.$$ $$axis$$ is line $$FS_1S_2.\ directrix$$ 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{align}

=&\ \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{align}$$

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 $$= \frac{2}{3}(base)(height) = \frac{2}{3} (\frac{2375}{144}) (\frac{361}{24}) = \frac{(2375)(361)}{3(12^3)}$$.