Introduction to Ellipse

DEFINITION

An ellipse is the locus of a point which moves in a plane such that the ratio of its distances from a fixed point (called focus) and from a fixed straight line (called directrix) is always constant and less than 1. And this constant ratio is called the eccentricity of the ellipse.

STANDARD EQUATION OF THE ELLIPSE

$\frac{x^2}{a^2}$ + $\frac{y^2}{a^2\left(1-e^2\right)}$ = 1 or $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,

where b² = a²(1 - e²).

The eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by the relation b² = a²(1 - e²), i.e., e² = 1 - b²/a²

An ellipse has two foci and two directrices.

Latus Rectum: Latus rectum is the line which passes through the focus of the ellipse and perpendicular to the major axis. End points of the latus rectum are given by(ae, b²/a) and (ae,-b²/a).Also the length of semi latus rectum is given by b² / a.

Focal Distance of a Point: To find the distance of any point on the ellipse from the focus, we use the definition of the ellipse.

Let P(x, y) be a point on the ellipse. Then

S'P = ePN' = e(a/e –x) = a - ex

SP = ePN = e ( a/e + x) = a + ex

S'P + SP = 2a

$\Rightarrow$ the sum of the focal distances of any point on the ellipse is equal to its major axis. Also SS'<SP+S'P=2a.

Other Forms of the Ellipse:

(i) If in the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, a²< b² , then the major and minor axis of the ellipse lie along the y and x –axis and are of lengths 2b and 2a respectively. The foci become (0, $\pm$ be) , and the directrices become y = $\pm$ b/e

where e = $\sqrt{1-\frac{a^2}{b^2}}$. The length of the semi-latus rectum becomes $\frac{a^2}{b}$.

(ii) If the centre of the ellipse be taken (h, k) and axes parallel to x and y-axes, then the equation of the ellipse is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$.

Illustration 1: Find the equation of ellipse referred to its centre whose foci are the points (4, 0) and (–4, 0) and whose eccentricity is 1/3.

Solution: Let the equation to the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ……(1)

Distance between the foci = 2ae = 4 + 4 = 8 ……(2)

Putting the value of e = 1/3 in (2), a = 4/e = 12

Again b² = a²(1 –e²) = 144 (1–1/9) = 128

Put in (1) we have $\frac{x^2}{144}+\frac{y^2}{128}=1$

$\Rightarrow$ 8x² + 9y² = 1152

GENERAL EQUATION OF THE ELLIPSE

Let the equation of the directrix of an ellipse be

ax + by + c = 0 and the focus be (h, k).

Let the eccentricity of the ellipse be e(e < 1).

If P(x, y) is any point on the ellipse, then

PS² = e² PM²

$\Rightarrow$ (x - h)² + (y - k)² = e²$\frac{{\left(ax+by+c\right)}^2}{a^2+b^2}$

$\Rightarrow$ which is of the form

$\Rightarrow$ ax²+2hxy +by²+2gx+2fy+c= 0, ... (1) where

$\Delta$ = abc +2 fgh –af²-bg² – ch² ¹ 0, h² < ab ,

Which are the necessary and sufficient condition for a general quadratic equation given by (1) to represent an ellipse .

Example -2: Find the eccentricity and latus rectum of the ellipse 4x² + 9y² – 8x – 36y + 4 = 0, also find centre, focus and directrix.

Solution: On simplification, equation becomes

4(x – 1)² + 9(y – 2)² = 36

Then, $\frac{X^2}{3^2}+\frac{Y^2}{2^2}=1$, where X = x – 1 and Y = y – 2, a = 3, b = 2

Now, e = $\sqrt{\frac{a^2-b^2}{a^2}}=\frac{\sqrt{5}}{3}$

The latus rectum = $\frac{2b^2}{a}=\frac{8}{3}$

For a > b

Focus (X, Y) is (ae, 0) and (–ae, 0)

$\Rightarrow$ (x, y) for focus $\Rightarrow$ (1 + ae, 2) and (1 – ae, 2)

$\Rightarrow$ foci are (1 + $\sqrt{5}$, 2) and (1 – $\sqrt{5}$, 2)

Similarly equation of directrices are X = $\frac{a}{e}$ and X = –$\frac{a}{e}$

$\Rightarrow$ x = 1 + $\frac{9}{\sqrt{5}}$ and x = 1 – $\frac{9}{\sqrt{5}}$ are the directrices.

AUXILIARY CIRCLE

The circle described on the major axis of an ellipse as diameter is called the Auxiliary Circle of the ellipse. Equation of the auxiliary circle is x²+y²=a².

PARAMETRIC EQUATION OF THE ELLIPSE

Thus for any point P(x, y) on the ellipse we have x = a cos$\theta$, y = b sin$\theta$ are called parametric equation of the ellipse.

Position of a Point Relative to an Ellipse:

The point P(x₁, y₁) is outside or inside or on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ according as the quantity

S₁ º $\left(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\right)$ is positive or negative or zero.

Illustration 3: Consider the ellipse x² + 3y² = 6 and a point P on it in the first quadrant at a distance of 2 units from the centre. Find the eccentric angle of P.

Solution: Equation of ellipse is x² + 3y² = 6

Equation of auxiliary circle is x² + y² = 6

Since P $\equiv$ (x₁, y₁) & Q $\equiv$ (x₁, y₂) lie on the ellipse and the circle respectively we have,

x₁² + 3y₁² = 6 …..(1)

x₁² + y₂² = 6 …..(2)

$\therefore$ 3y₁² –y₂² = 0 $\Rightarrow$ $\frac{y_2}{y_1}=\sqrt{3}$

Again OP = 2 $\Rightarrow$ x₁² + y₁² = 4 …..(3)

By (1) –(3), we get,

2y₁² = 2 $\Rightarrow$ y₁² = 1 $\Rightarrow$ y₁ = 1 [$\because$ P is in the first quadrant]

$\therefore$y₂ = $\sqrt{3}$

Putting y₁ in (1), we get x₁² = 3 $\Rightarrow$ x₁ = $\sqrt{3}$

$\therefore$ Eccentric angle of P = $\theta$ = ${\tan }^{-1}\left(\frac{y_2}{x_1}\right)={\tan }^{-1}\left(\frac{\sqrt{3}}{\sqrt{3}}\right)=\frac{\pi }{4}$