# Question 4f402

Jul 19, 2016

Reqd. Eqn. of the Ellipse is $: {x}^{2} / 12 + {y}^{2} / 16 = 1$. graph{x^2/12+y^2/16=1 [-10, 10, -5, 5]}

#### Explanation:

Let us suppose that the reqd. eqn. of ellipse is

$S : {x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$.

Given that pt.$\left(- 3 , 2\right) \in S \Rightarrow \frac{9}{a} ^ 2 + \frac{4}{b} ^ 2 = 1. \ldots \ldots \ldots . . \left(1\right)$.

We take a note, that the Major Axis of the Ellipse is Y-axis.

$\therefore b > a , \mathmr{and} , {a}^{2} = {b}^{2} \left(1 - {e}^{2}\right) \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right) \left(e < 1\right)$.

Length $l$ of Latus Retum$= 2 {a}^{2} / b$

Centre is $C \left(0 , 0\right)$ and Focii are $S \left(0 , b e\right) \mathmr{and} S \left(0 - b e\right)$, so that,

$C S , \mathmr{and} , C S ' = b e$.

By what is given, $l = 3 C S \Rightarrow 2 {a}^{2} / b = 3 b e \Rightarrow 2 {a}^{2} = 3 {b}^{2} e$

By $\left(2\right)$, then, $2 {\cancel{b}}^{2} \left(1 - {e}^{2}\right) = 3 {\cancel{b}}^{2} e$

$\therefore 2 {e}^{2} + 3e-2 = 0 \Rightarrow \left(e + 2\right) \left(2e-1\right) = 0$, giving,

$e = - 2$, which is impossible, or, e=1/2(<1,# as desired.)

Then, from $\left(2\right) , {a}^{2} = \frac{3}{4} {b}^{2}$, or, ${a}^{2} / 3 = {b}^{2} / 4 , i . e . , \frac{4}{b} ^ 2 = \frac{3}{a} ^ 2$

Using this in $\left(1\right) , \frac{9}{a} ^ 2 + \frac{3}{a} ^ 2 = 1 , i . e . , \frac{12}{a} ^ 2 = 1 , \mathmr{and} , {a}^{2} = 12$

Finally, since, ${b}^{2} = \frac{4}{3} {a}^{2} = \frac{4}{3} \cdot 12 = 16$

Thus, with ${a}^{2} = 12 , \mathmr{and} , {b}^{2} = 16$ we get the reqd. eqn. of Ellipse,

$S : {x}^{2} / 12 + {y}^{2} / 16 = 1$.