Parabola

File:20170525 Parabola y = (x^2)-(4q) 00.png

Figure 1: The Parabola $y = \frac{x^{2}}{4 q}$
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} .

V F = V D_{3} = q .

In Figure 1

q = 1

In Cartesian geometry in two dimensions the $p a r a b o l a$ 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 $f o c u s$ and the fixed line is called the $d i r e c t r i x$ . Distance from $f o c u s$ to $d i r e c t r i x$ 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 $v e r t e x$ , point $V$ $: (0, 0)$ , is half-way between focus and directrix. A $c h o r d$ is the segment of a line joining any two distinct points of the parabola. The line segment $P_{2} F P_{4}$ is a chord. Because chord $P_{2} F P_{4}$ passes through the focus $F$ , it is called a $f o c a l c h o r d .$

The focal chord parallel to the directrix is called the $l a t u s r e c t u m .$

The line through the focus and perpendicular to the directrix is the $a x i s$ , sometimes called $a x i s o f s y m m e t r y$ .

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

.

If the equation of the curve is expressed as: $y = K x^{2}$ , then $K = \frac{1}{4 q}; 4 K q = 1; q = \frac{1}{4 K}$ where the $f o c u s$ 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 $a x + b y + c = 0$ 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 ( $a x + b y + c = 0$ )

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

By definition these two lengths are equal:

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

$∴ \sqrt{R} \sqrt{(x - p)^{2} + (y - q)^{2}} = a x + b y + c$ .

Square both sides:

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

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

Expand and the result is:

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

$(1)$ has the form of the equation of the conic section $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ where

$\begin{matrix} A & = b^{2} \\ B & = - 2 a b \\ C & = a^{2} \\ D & = - 2 (a c + p R) \\ E & = - 2 (b c + q R) \\ F & = R (p^{2} + q^{2}) - c^{2} \\ B & ^{2} - 4 A C = 0 \\ R & = a^{2} + b^{2} \end{matrix}$

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

An Example

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File:20170525 General parabola 00.png

Figure 2: The Parabola $16 x^{2} + 24 x y + 9 y^{2} - 20 x - 140 y + 100 = 0$
Green line is

d i r e c t r i x : 3 x - 4 y - 5 = 0

Blue line is

a x i s : 4 x + 3 y - 10 = 0

Focus at point

F : (1, 2)

Vertex at point

V : (1.6, 1.2)

V F = 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:

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

The equation of the parabola in Figure 2 is: $16 x^{2} + 24 x y + 9 y^{2} - 20 x - 140 y + 100 = 0 .$

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

Distance from $f o c u s$ to $d i r e c t r i x = \frac{3}{5} (1) - \frac{4}{5} (2) - 1 = - 2 .$

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

Therefore, curve has shape of $y = K x^{2}$ where $K = \frac{1}{4}$ . Template:RoundBoxTop Caution: An interesting situation occurs if the focus is on the directrix. Consider the directrix:

$4 x - 3 y + 15 = 0$ 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: $9 x x + 24 x y + 16 y y - 270 x - 360 y + 2025 = 0$ .

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

$9 x x + 24 x y + 16 y y - 270 x - 360 y + 2025 = (3 x + 4 y - 45)^{2}$ .

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 $= \frac{4}{5} (3) + \frac{- 3}{5} (9) + 3 = 0 .$ Template:RoundBoxBottom Template:RoundBoxBottom

Reverse-Engineering the Parabola

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File:20170526 Parabola and 2 tangents 00.png

Figure 3: Parabola with 2 tangents parallel to axes.
Tangent

A B C : y = \frac{16}{3}

.
Tangent

A D E : x = - 0.25

.
Point

A

on directrix, oblique, thin, black line.

Line

A F G

perpendicular to focal chord

B F D .

Focus at

F : (1, 7)

.

Given a parabola in form $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ the aim is to produce the directrix and the focus.

We will solve the example shown in Figure 3: $9 x^{2} - 24 x y + 16 y^{2} + 70 x - 260 y + 1025 = 0$ ,

where:

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

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

Method 1. By analytical geometry

Template:RoundBoxTop Find two tangents that intersect at a right angle. The simplest to find are those that are parallel to the axes. Template:RoundBoxTop Put the equation of the parabola in the form of a quadratic in $x$ .

$A x^{2} + B x y + D x + C y^{2} + E y + F = 0$

$A x^{2} + (B y + D) x + (C y^{2} + E y + F) = 0$

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

$(B y + D)^{2} - 4 (A) (C y^{2} + E y + F) = 0$

$B B y y + 2 B D y + D D - 4 A C y y - 4 A E y - 4 A F = 0$

$(B B - 4 A C) y y + (2 B D - 4 A E) y + (D D - 4 A F) = 0$

In the general parabola $B^{2} - 4 A C = 0$ therefore $y = \frac{4 A F - D D}{2 B D - 4 A E}$ .

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

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

The line $A B C$ is tangent to the curve at $B$ and has equation: $y = \frac{16}{3}$ . Template:RoundBoxBottom Template:RoundBoxTop Put the equation of the parabola in the form of a quadratic in $y$ :

$C y^{2} + (B x + E) y + (A x^{2} + D x + F) = 0$ .

By using calculations similar to the above, $x = \frac{4 C F - E E}{2 B E - 4 C D} = - 0.25$ and $y = \frac{- (B x + E)}{2 C} = 7.9375$ .

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

The line $A D E$ is tangent to the curve at $D$ and has equation: $x = - 0.25$ . Template:RoundBoxBottom Template:RoundBoxTop 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: $a x + b y + c = 0$ .

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: $4 x + 3 y - 15 = 0$ . Template:RoundBoxBottom Template:RoundBoxTop The coordinates of points $B, D$ are known. Therefore chord $B D$ is defined as: $\frac{3}{5} x + \frac{4}{5} y - 6.2 = 0$ .

Draw the line $A G$ perpendicular to $B D$ . The line $A G$ is defined as: $\frac{4}{5} x - \frac{3}{5} y + 3.4 = 0$ .

Point $F$ at the intersection of lines $B D, A G$ is the focus with coordinates $(1, 7)$ . Template:RoundBoxBottom Template:RoundBoxBottom

Method 2. By algebra

Template:RoundBoxTop

\begin{matrix} A & = b^{2} \\ B & = - 2 a b \\ C & = a^{2} \\ D & = - 2 (a c + p R) \\ E & = - 2 (b c + q R) \\ F & = R (p^{2} + q^{2}) - c^{2} \\ B & ^{2} - 4 A C = 0 \\ R & = a^{2} + b^{2} \end{matrix}

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

\begin{matrix} D + 2 a c + 2 p R = 0 \\ E + 2 b c + 2 q R = 0 \\ F - R p p - R q q + c c = 0 \end{matrix}

The solutions are:

$c = \frac{4 F R - (D D + E E)}{4 D a + 4 E b}$

$p = \frac{- (D + 2 a c)}{2 R}$

$q = \frac{- (E + 2 b c)}{2 R}$

Template:RoundBoxTop If $b$ is $0, A = B = 0, R = a^{2} = C,$ the parabola becomes the quadratic: $- D x = C y^{2} + E y + F$ and:

\begin{matrix} p & = \frac{E^{2} - D^{2} - 4 F C}{4 C D} \\ q & = \frac{- E}{2 C} \\ c & = \frac{4 F C - (D D + E E)}{4 D a} \end{matrix}

The directrix has equation: $a x + c = 0; a x = - c; x = \frac{D D + E E - 4 F C}{4 C D}$ .

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

\begin{matrix} x & = C y^{2} + E y + F \\ p & = \frac{E^{2} - 1 - 4 F C}{4 C (- 1)} = \frac{1 - (E^{2} - 4 C F)}{4 C} \\ q & = \frac{- E}{2 C} \end{matrix}

The directrix has equation: $x = \frac{- 1 - (E^{2} - 4 C F)}{4 C}$ . Template:RoundBoxBottom

Template:RoundBoxTop If $a$ is $0, B = C = 0, R = b^{2} = A,$ the parabola becomes the quadratic: $- E y = A x^{2} + D x + F$ and:

\begin{matrix} p & = \frac{- D}{2 A} \\ q & = \frac{D^{2} - E^{2} - 4 F A}{4 E A} \\ c & = \frac{4 F A - (D D + E E)}{4 E b} \end{matrix}

The directrix has equation: $b y + c = 0; b y = - c; y = \frac{D D + E E - 4 F A}{4 E A}$ .

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

\begin{matrix} y & = A x^{2} + D x + F \\ p & = \frac{- D}{2 A} \\ q & = \frac{D^{2} - 1 - 4 F A}{4 (- 1) A} = \frac{1 - (D^{2} - 4 A F)}{4 A} \end{matrix}

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

These values agree with the corresponding values in the graph of $y = x^{2} - 2 x - 3$ to the right. Template:RoundBoxBottom Template:RoundBoxBottom Template:RoundBoxBottom

Slope of the Parabola

Consider parabola $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ and line $y = m x + c .$

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

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

\begin{matrix} (+ A + B m + C m m) & x x + \\ (- B X m + B Y - 2 C X m m + 2 C Y m + D + E m) & x + \\ (+ C X X m m - 2 C X Y m + C Y Y - E X m + E Y + F) & = 0 \end{matrix}

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

Discriminant =

\begin{matrix} (- B X m + B Y - 2 C X m m + 2 C Y m + D + E m) (- B X m + B Y - 2 C X m m + 2 C Y m + D + E m) \\ - 4 (+ A + B m + C m m) (+ C X X m m - 2 C X Y m + C Y Y - E X m + E Y + F) & = 0 \end{matrix}

The above discriminant is a quadratic in $m$ :

\begin{matrix} (- 4 A C X X + B B X X + 2 B E X - 4 C D X - 4 C F + E E) & m m + \\ (+ 8 A C X Y + 4 A E X - 2 B B X Y - 2 B D X - 2 B E Y - 4 B F + 4 C D Y + 2 D E) & m + \\ (- 4 A C Y Y - 4 A E Y - 4 A F + B B Y Y + 2 B D Y + D D) & = 0 \end{matrix}

$m = \frac{- (+ 8 A C X Y + 4 A E X - 2 B B X Y - 2 B D X - 2 B E Y - 4 B F + 4 C D Y + 2 D E) \pm \sqrt{R}}{2 (- 4 A C X X + B B X X + 2 B E X - 4 C D X - 4 C F + E E)}$

where:

\begin{matrix} R = & 16 (A X X + B X Y + C Y Y + D X + E Y + F) (- 4 A C F + A E E + B B F - B D E + C D D) \end{matrix}

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

\begin{matrix} (A X & X + B X Y + C Y Y + D X + E Y + F) = 0; R = 0 \\ m = & \frac{- (+ 8 A C X Y + 4 A E X - 2 B B X Y - 2 B D X - 2 B E Y - 4 B F + 4 C D Y + 2 D E)}{2 (- 4 A C X X + B B X X + 2 B E X - 4 C D X - 4 C F + E E)} \\ = & \frac{2 B B X Y - 8 A C X Y + 2 B D X - 4 A E X + 2 B E Y - 4 C D Y + 4 B F - 2 D E}{2 (B B X X - 4 A C X X + 2 B E X - 4 C D X + E E - 4 C F)} \\ = & \frac{(B B - 4 A C) X Y + (B D - 2 A E) X + (B E - 2 C D) Y + 2 B F - D E}{(B B - 4 A C) X X + (2 B E - 4 C D) X + E E - 4 C F} \\ = & \frac{(B D - 2 A E) X + (B E - 2 C D) Y + 2 B F - D E}{(2 B E - 4 C D) X + E E - 4 C F} \end{matrix}

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

When slope is displayed in this format, we see that slope is vertical if $(2 B E - 4 C D) X + E E - 4 C F = 0 .$

The line $x = \frac{4 C F - E E}{2 B E - 4 C D}$ is tangent to the curve. Template:RoundBoxTop Let the equation of a line be: $x = M y + c$ in which case $M = \frac{1}{m} .$

By using calculations similar to the above it can be shown that:

$M = \frac{(B D - 2 A E) X + (B E - 2 C D) Y + 2 B F - D E}{(2 B D - 4 A E) Y + D D - 4 A F}$ .

$m = \frac{1}{M}$ therefore:

$m = \frac{(2 B D - 4 A E) Y + D D - 4 A F}{(B D - 2 A E) X + (B E - 2 C D) Y + 2 B F - D E}$

When slope is displayed in this format, we see that slope is zero if $(2 B D - 4 A E) Y + D D - 4 A F = 0$ .

The line $y = \frac{4 A F - D D}{2 B D - 4 A E}$ is tangent to the curve. Template:RoundBoxBottom Template:RoundBoxTop This examination of the parabola has produced two expressions for slope of the parabola:

$m = \frac{(2 B D - 4 A E) y + D D - 4 A F}{(B D - 2 A E) x + (B E - 2 C D) y + 2 B F - D E} = \frac{(B D - 2 A E) x + (B E - 2 C D) y + 2 B F - D E}{(2 B E - 4 C D) x + E E - 4 C F}$ .

where the point $(x, y)$ is any arbitrary point on the curve.

Therefore $m = \pm \sqrt{\frac{(2 B D - 4 A E) y + D D - 4 A F}{(2 B E - 4 C D) x + E E - 4 C F}}$ . This formula for $m$ contains both tangents parallel to the axes and is derived without calculus. Template:RoundBoxBottom Template:RoundBoxTop

The formula from calculus below is simpler and unambiguous concerning sign.

\begin{matrix} A x^{2} + B x y + C y^{2} + D x + E y + F = 0 \\ \frac{d}{d x} (A x^{2} + B x y + C y^{2} + D x + E y + F) = 0 \\ A (2 x) + B (x \frac{d y}{d x} + y \frac{d x}{d x}) + C (2 y \frac{d y}{d x}) + D + E (\frac{d y}{d x}) = 0 \\ A (2 x) + B (x \frac{d y}{d x} + y) + C (2 y \frac{d y}{d x}) + D + E (\frac{d y}{d x}) = 0 \\ 2 A x + B x \frac{d y}{d x} + B y + 2 C y \frac{d y}{d x} + D + E \frac{d y}{d x} = 0 \\ \frac{d y}{d x} (B x + 2 C y + E) = - (2 A x + B y + D) \\ m = \frac{d y}{d x} = \frac{- (2 A x + B y + D)}{B x + 2 C y + E} \end{matrix}

Template:RoundBoxBottom Template:RoundBoxTop The slope of the parabola is $0$ where $(2 B D - 4 A E) y + D D - 4 A F = 0; 2 A x + B y + D = 0;$ or:

$y = \frac{4 A F - D D}{2 B D - 4 A E}; x = \frac{- (B y + D)}{2 A} .$

The slope of the parabola is vertical where $(2 B E - 4 C D) x + E E - 4 C F = 0; B x + 2 C y + E = 0;$ or:

$x = \frac{4 C F - E E}{2 B E - 4 C D}; y = \frac{- (B x + E)}{2 C} .$ Template:RoundBoxBottom Template:RoundBoxTop Caution:

If the curve is $y = K x^{2}$ , the slope can never be vertical.

If the curve is $x = K y^{2}$ , the slope can never be $0$ . Template:RoundBoxBottom Template:RoundBoxTop If $B = C = 0$ and $E = - 1$ , the equation of the parabola becomes: $y = A x^{2} + D x + F$ and:

\begin{matrix} m = & \frac{(2 B D - 4 A E) y + D D - 4 A F}{(B D - 2 A E) x + (B E - 2 C D) y + 2 B F - D E} \\ = & \frac{(- 4 A (- 1)) y + D D - 4 A F}{(- 2 A (- 1)) x - D (- 1)} = \frac{4 A y + D D - 4 A F}{2 A x + D} \\ = & \frac{4 A (A x^{2} + D x + F) + D D - 4 A F}{2 A x + D} = \frac{4 A^{2} x^{2} + 4 A D x + 4 A F + D D - 4 A F}{2 A x + D} \\ = & \frac{4 A^{2} x^{2} + 4 A D x + D D}{2 A x + D} \\ = & 2 A x + D \\ m = & \frac{(B D - 2 A E) x + (B E - 2 C D) y + 2 B F - D E}{(2 B E - 4 C D) x + E E - 4 C F} \\ = & \frac{((0) D - 2 A (- 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{(- 2 A (- 1)) x - D (- 1)}{(- 1) (- 1)} \\ = & 2 A x + D \\ m = & \frac{- (2 A x + B y + D)}{B x + 2 C y + E} \\ = & \frac{- (2 A x + (0) y + D)}{(0) x + 2 (0) y + (- 1)} \\ = & 2 A x + D \end{matrix}

Template:RoundBoxBottom

Point at given slope

Template:RoundBoxTop Given parabola defined by $A, B, C, D, E, F$ and slope $m = \frac{y v a l u e}{x v a l u e}$ where at least one of $y v a l u e, x v a l u e$ is non-zero, calculate point at which the slope is $m .$

Let $m = \frac{- (2 A x + B y + D)}{B x + 2 C y + E} = \frac{s}{t} = \frac{y v a l u e}{x v a l u e}$

Then $s (B x + 2 C y + E) + t (2 A x + B y + D) = 0 .$

Let $G = (B s + 2 A t); H = (2 C s + B t); I = (E s + D t) .$

Then $G x + H y + I = 0 \dots (1)$

Substitute in the equation of the parabola and $y = \frac{- (A I I - D G I + F G G)}{(2 A H I - B G I - D G H + E G G)} \dots (2)$

As shown below, with a little manipulation of the data, the same formula can be used to calculate $x .$ Template:RoundBoxTop 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. Template:RoundBoxBottom Template:RoundBoxBottom

Implementation

Template:RoundBoxTop

# 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

Template:RoundBoxBottom

Examples

Template:RoundBoxTop A parabola is defined as $9 x^{2} - 24 x y + 16 y^{2} + 70 x - 260 y + 1025 = 0 .$

Template:RoundBoxTop 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)

Template:RoundBoxBottom Template:RoundBoxTop 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)

Template:RoundBoxBottom Template:RoundBoxBottom

Parabola and any chord

File:20170418ParabolaWithChord&2tangents.pdf

Figure 4: Parabola and any chord.
Origin

(0, 0)

at point

O;

curve

:

y = K x^{2};

chord

I J

: y = m x + c;

point

L

: (0, c);

2 tangents

:

I N, J N;

point

N

: (\frac{m}{2 K}, - c)

Refer to Figure 4.

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

The chord $I J$ has equation: $y = m x + c$ .

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

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

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

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

\begin{matrix} y = & K x^{2} = m x + c \\ K x^{2} & - m x - c = 0 \\ x = & \frac{m \pm \sqrt{m^{2} + 4 K c}}{2 K} \\ = & \frac{m \pm R}{2 K} \end{matrix}

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

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

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

$y_{1} = K x_{1}^{2} = K (\frac{m - R}{2 K}) (\frac{m - R}{2 K}) = \frac{m^{2} - 2 m R + R^{2}}{4 K} = \frac{m^{2} - m R + 2 K c}{2 K}$ ,

and slope of tangent $I N = s_{1} = 2 K x_{1} = m - R .$

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

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

$y_{2} = \frac{m^{2} + m R + 2 K c}{2 K}$ ,

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

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

Equation of tangent $I N :$

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

Equation of tangent $J N :$

\begin{matrix} y = (m + R) x - \frac{m^{2} + m R + 2 K c}{2 K} \end{matrix}

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

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

The line $N G H$ with equation $x = \frac{m}{2 K}$ bisects the line segment $D E$ and also the chord $I J$ at point $H$ .

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

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

The $Y$ coordinate of point $N :$

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

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

Angle $I N J$ Template:RoundBoxTop Using:

\begin{matrix} \tan (A - B) = & \frac{\tan (A) - \tan (B)}{1 + \tan (A) \tan (B)} \\ \tan (∠ I N J) = & \frac{(m - R) - (m + R)}{1 + (m - R) (m + R)} \\ = & \frac{- 2 R}{1 + (m^{2} - R^{2})} \\ = & \frac{- 2 R}{1 + m^{2} - (m^{2} + 4 K c)} \\ = & \frac{- 2 R}{1 - 4 K c} \\ = & \frac{2 \sqrt{m^{2} + 4 K c}}{4 K c - 1} \end{matrix}

If $4 K c > 1$ , point $L$ is above the $f o c u s, \tan (∠ I N J)$ is positive and $0$ ° $< ∠ I N J < 90$ °.

$∠ I N J$ is acute and, as $m$ increases, $∠ I N J$ increases, approaching $90$ °.

If $4 K c = = 1$ , point $L$ is on the $f o c u s, \tan (∠ I N J) = \frac{2 \sqrt{m^{2} + 1}}{0}, ∠ I N J = 90$ ° and the line $y = - c$ is the $d i r e c t r i x$ .

If $4 K c < 1$ , point $L$ is below the $f o c u s, \tan (∠ I N J)$ is negative and $90$ ° $< ∠ I N J < 180$ °.

$∠ I N J$ is obtuse and, as $m$ increases, $∠ I N J$ decreases, approaching $90$ °. Template:RoundBoxBottom

Parabola and two tangents

File:20170418ParabolaWithChord&2tangents.pdf

Figure 5: Parabola and two tangents.
Origin

(0, 0)

at point

O;

curve

:

y = K x^{2};

point

N

: (h, - c);

2 tangents

:

N I, N J;

chord

I J

: y = 2 K h x + c;

point

L

: (0, c);

Refer to Figure 5.

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

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

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

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

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

Equation of any line through point $N : y = m x - c - m h$

Let this line intersect the curve:

$y = K x^{2} = m x - c - m h$

$K x^{2} - m x + c + m h = 0$

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

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

\begin{matrix} m^{2} - & 4 K m h - 4 K c = 0 \\ m = & \frac{4 K h \pm \sqrt{(4 K h)^{2} - 4 (1) (- 4 K c)}}{2} \\ = & \frac{4 K h \pm \sqrt{16 K^{2} h^{2} + 16 K c}}{2} \\ = & 2 K h \pm 2 R \end{matrix}

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

$m_{1} = 2 K h - 2 R =$ slope of tangent $N I$ .

$m_{2} = 2 K h + 2 R =$ slope of tangent $N J$ .

\begin{matrix} x = & \frac{m}{2 K} \\ x_{1} = & \frac{m_{1}}{2 K} = \frac{2 K h - 2 R}{2 K} = \frac{K h - R}{K} \\ y_{1} = & K x_{1}^{2} = K (\frac{K h - R}{K}) (\frac{K h - R}{K}) = \frac{(K h - R) (K h - R)}{K} \\ x_{2} = & \frac{K h + R}{K} \\ y_{2} = & \frac{(K h + R) (K h + R)}{K} \end{matrix}

Slope of chord $I J$

\begin{matrix} = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ = & (\frac{(K h + R) (K h + R)}{K} - \frac{(K h - R) (K h - R)}{K}) / (\frac{K h + R}{K} - \frac{K h - R}{K}) \\ = & (\frac{K K h h + 2 K h R + R R - (K K h h - 2 K h R + R R)}{K}) / (\frac{2 R}{K}) \\ = & (\frac{4 K h R}{K}) (\frac{K}{2 R}) \\ = & 2 K h \end{matrix}

Intercept of chord $I J$ on the $Y$ axis

\begin{matrix} = & y_{1} - 2 K h x_{1} \\ = & \frac{(K h - R) (K h - R)}{K} - 2 K h \frac{K h - R}{K} \\ = & \frac{K K h h - 2 K h R + R R}{K} - \frac{2 K h (K h - R)}{K} \\ = & \frac{K K h h - 2 K h R + R R - 2 K h K h + 2 K h R}{K} \\ = & \frac{K K h h + R R - 2 K h K h}{K} \\ = & \frac{- K K h h + K K h h + K c}{K} \\ = & \frac{+ K c}{K} \\ = & c \end{matrix}

Angle $I N J$

If $∠ I N J = = 90$ °

\begin{matrix} (m_{1}) (m_{2}) = - 1 \\ (2 K h - 2 R) (2 K h + 2 R) = - 1 \\ 4 (K K h h - R R) = - 1 \\ 4 (K K h h - (K K h h + K c)) = - 1 \\ 4 (- K c) = - 1 \\ 4 K c = 1 \end{matrix}

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

$y = K x^{2} ∴ K = \frac{1}{4 q}$ or $4 K q = 1 .$

If $4 K c = = 1,$ then $(0, c)$ and $(0, q)$ are the same point, the chord $I J$ passes through the $f o c u s$ and the line $y = - c$ is the $d i r e c t r i x$ .

Area enclosed between parabola and chord

Introduction

Template:RoundBoxTop

File:20170526 Parabola and 2 chords 00.png

Figure 6: The Parabola: $y = x^{2}$
Chord

D C

, parallel tangent

A O B

and area

D O C D .

b a s e = C D = 2 . h e i g h t = H_{1} T_{1} = 1 .

Chord

O C

, parallel tangent

G H I

and area

O H C O

.

b a s e = O C = \sqrt{2} . h e i g h t = 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 $(O B C)$

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

Area under curve $(D O C) =$ area $(O A D) +$ area $(O B C) = \frac{2}{3}$

Area between chord $D C$ and curve $D O C = 2 - \frac{2}{3} = \frac{4}{3}$ .

Consider the chord $C D$ . Call this the $b a s e$ with value $2$ . The tangent $A O B$ through the origin is parallel to $b a s e (D C)$ , and the perpendicular distance between $A B, D C (H_{1} T_{1})$ is $1$ . Call this distance the $h e i g h t$ with value $1$ .

In this case the area enclosed between chord $D C$ and curve $D O C = \frac{2}{3} (b a s e) (h e i g h t) = \frac{2}{3} (2) (1) = \frac{4}{3},$ the same as that calculated earlier.

Consider the chord $O C$ and curve $O H C .$ By inspection the area between chord $O C$ and curve $O H C = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} .$

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

The line $G H I$ is parallel to $b a s e O C$ and touches the curve at $H (\frac{1}{2}, \frac{1}{4}) .$

Distance from $H$ to chord $O C = H_{2} T_{2} = \frac{1 / 2}{\sqrt{2}} - \frac{1 / 4}{\sqrt{2}} = \frac{1}{4 \sqrt{2}} = \frac{\sqrt{2}}{8} = h e i g h t .$

Length of $O C = \sqrt{2} = b a s e .$

Area between chord $O C$ and curve $O H C = \frac{2}{3} (b a s e) (h e i g h t) = \frac{2}{3} (\sqrt{2}) (\frac{\sqrt{2}}{8}) = \frac{2}{3} (\frac{1}{4}) = \frac{1}{6},$ the same as that calculated earlier.

These observations suggest that the area enclosed between chord and curve $= \frac{2}{3} (b a s e) (h e i g h t) .$ Template:RoundBoxBottom

Proof

Template:RoundBoxTop

File:20170527 Parabola & area under chord 00.png

Figure 7: The Parabola: $y = K x^{2}$
Origin at point

O : (0, 0)

Points

D, I : (p, 0), (p, K p^{2})

Points

E, J : (q, 0), (q, K q^{2})

Chord

I J

, parallel tangent

F G

and area

I O J I

, the area enclosed between chord

I J

and curve.
Length

I J = b a s e .

Length

H T = h e i g h t .

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

Slope of chord $I J = \frac{K q q - K p p}{q - p} = \frac{K ((q + p) (q - p))}{q - p} = K (q + p) .$

Find equation of chord $I J .$ $y = K (p + q) x + c; ∴ c = y - K (p + q) x = K q q - K (p + q) q = - K p q$

Equation of chord $I J :$ $y = K (p + q) x - K p q .$

Find equation of tangent $F G .$

\begin{matrix} y = K x^{2} \\ y = K (p + q) x + c \\ ∴ & K x^{2} = K (p + q) x + c \\ K x^{2} - K (p + q) x - c = 0 \end{matrix}

We choose a value of $c$ that gives $x$ exactly one value.

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

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

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

Therefore distance between chord $I J$ and tangent $F G$

$= h e i g h t = \frac{- K p q - (- \frac{K (p + q)^{2}}{4})}{R} = \frac{K (p + q)^{2} - 4 K p q}{4 R} = \frac{K (p - q)^{2}}{4 R}$ where $R = \sqrt{K^{2} (p + q)^{2} + 1} .$

Length of chord $I J = b a s e = L = \sqrt{(K q q - K p p)^{2} + (q - p)^{2}} .$

Area under chord $I J = (q - p) \frac{K q q + K p p}{2} = \frac{K q q q + K p p q - K q q p - K p p p}{2}$

Area under curve $I O J$

\begin{matrix} x = & q \\ = & [\frac{K x^{3}}{3}] = \frac{K q q q - K p p p}{3} \\ x = & p \end{matrix}

Area between chord $I J$ and curve $I O J$

$= \frac{K q q q + K p p q - K q q p - K p p p}{2} - \frac{K q q q - K p p p}{3} = \frac{K q q q + 3 K p p q - 3 K p q q - K p p p}{6}$

$= \frac{K (q - p)^{3}}{6} = \frac{K S}{6} .$

The aim is to prove that:

$\frac{2}{3} (b a s e) (h e i g h t) = \frac{K S}{6}$ or

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

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

$L (p - q)^{2} = R (q - p)^{3}$ or

$L = R (q - p)$

where:

$L = \sqrt{(K q q - K p p)^{2} + (q - p)^{2}} = \sqrt{(q - p)^{2} (K^{2} (q + p)^{2} + 1)} = (q - p) \sqrt{K^{2} (q + p)^{2} + 1}$

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

$R H S = (q - p) R = (q - p) \sqrt{K^{2} (p + q)^{2} + 1} = L .$

Therefore $L = R (q - p)$ and area enclosed between curve and chord = $\frac{2}{3} (b a s e) (h e i g h t)$

where $b a s e$ is the length of the chord, and $h e i g h t$ is the perpendicular distance between chord and tangent parallel to chord. Template:RoundBoxBottom

A worked example

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

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

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

Template:RoundBoxTop

File:20170526 Parabola and Chord 00.png

Figure 8: The Parabola: $16 x^{2} - 24 x y + 9 y^{2} + 20 x - 140 y + 600 = 0$
Chord

A D : 4 x + 3 y - 96 = 0 .

Length

A D = b a s e = \frac{2375}{144} .

Parallel tangent

C F B . h e i g h t = B E = \frac{361}{24} .

Area between chord and curve

= D E A F D .

Length of chord

\begin{matrix} = & \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} = b a s e . \end{matrix}

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 $g = \frac{4 A F e e - 4 B F d e + 4 C F d d - D D e e + 2 D E d e - E E d d}{4 A E e - 2 B D e - 2 B E d + 4 C D d}$

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 $= - \frac{499}{120} - (- 19.2) = \frac{361}{24} = h e i g h t .$

Area enclosed between chord and curve $= \frac{2}{3} (b a s e) (h e i g h t) = \frac{2}{3} (\frac{2375}{144}) (\frac{361}{24}) = \frac{(2375) (361)}{3 (1 2^{3})} .$ Template:RoundBoxBottom

Method 2. By identifying the basic parabola.

Template:RoundBoxTop

File:20170526 Determine basic parabola 00.png

Figure 9: The Parabola: $16 x^{2} - 24 x y + 9 y^{2} + 20 x - 140 y + 600 = 0$
Chord

A D : 4 x + 3 y - 96 = 0 .

a x i s

is line

F S_{1} S_{2} . d i r e c t r i x

is line

R T .

f o c u s

is point

F : (2, 6) .

Line

A S_{2} = p = - 8 \frac{1}{2} .

Line

D S_{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 $y = K x^{2}$ 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{matrix} = & \frac{K (q - p)^{3}}{6} \\ = & \frac{1}{24} (\frac{95}{6}) (\frac{95}{6}) (\frac{95}{6}) \\ = & \frac{(2375) (361)}{3 (1 2^{3})} . \end{matrix}

or:

$h e i g h t = \frac{K (p - q)^{2}}{4 R}$ where $R = \sqrt{K^{2} (p + q)^{2} + 1}$

$(p - q)^{2} = \frac{95 (95)}{36}$

$(p + q)^{2} = \frac{49}{36}$

$R = \sqrt{\frac{49}{16 (36)} + 1} = \sqrt{\frac{7^{2}}{2 4^{2}} + \frac{2 4^{2}}{2 4^{2}}} = \sqrt{\frac{2 5^{2}}{2 4^{2}}} = \frac{25}{24}$

$h e i g h t = \frac{\frac{95 (95)}{(4) 36}}{4 (\frac{25}{24})} = \frac{\frac{95 (95)}{(4) 36}}{\frac{25}{6}} = (\frac{95 (95)}{(4) 36}) (\frac{6}{25}) = \frac{19 (19)}{24} = \frac{361}{24} .$

Calculate $b a s e$ as above and area enclosed between chord and curve $= \frac{2}{3} (b a s e) (h e i g h t) = \frac{2}{3} (\frac{2375}{144}) (\frac{361}{24}) = \frac{(2375) (361)}{3 (1 2^{3})}$ . Template:RoundBoxBottom

Parabola

Contents

The General Parabola

An Example

Reverse-Engineering the Parabola

Method 1. By analytical geometry

Method 2. By algebra

Slope of the Parabola

Point at given slope

Implementation

Examples

Parabola and any chord

Parabola and two tangents

Area enclosed between parabola and chord

Introduction

Proof

A worked example

Method 1. By chord and parallel tangent

Method 2. By identifying the basic parabola.

Navigation menu

Parabola

The General Parabola

An Example

Reverse-Engineering the Parabola

Method 1. By analytical geometry

Method 2. By algebra

Slope of the Parabola

Point at given slope

Implementation

Examples

Parabola and any chord

Parabola and two tangents

Area enclosed between parabola and chord

Introduction

Proof

A worked example

Method 1. By chord and parallel tangent

Method 2. By identifying the basic parabola.

Navigation menu

Search