Question

Find the equation of the ellipse with

length of latus rectum 4, distance between foci 4 \sqrt{2} and major axis as y -axis

.

Accepted Answer

The given ellipse is having $length of latus rectum 4, distance between foci 4 \sqrt{2} and major axis as y -axis$ .

Here, the ellipse is along $y -axis$ . So, the ellipse is in form of $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 . . . (i)$

The distance between foci is $4 \sqrt{2} \Rightarrow 2 a e = 4 \sqrt{2} \Rightarrow a e = 2 \sqrt{2}$

The length of latus rectum is $4 \Rightarrow \frac{2 b^{2}}{a} = 4 \Rightarrow 2 b^{2} = 4 a$

$∴ b^{2} = 2 a$

By using formula of eccentricity of an ellipse, we have $e^{2} = \frac{a^{2} - 2 a}{a^{2}} \Rightarrow {(a e)}^{2} = a^{2} - 2 a \Rightarrow {(2 \sqrt{2})}^{2} = a^{2} - 2 a$

$\Rightarrow a^{2} - 2 a - 8 = 0 \Rightarrow a^{2} + 2 a - 4 a - 8 = 0 \Rightarrow (a + 2) (a - 4) = 0$

$∴ a = 4 [∵ b^{2} can't be negative]$

$Now, a = 4 and b^{2} = 2 \times 4 = 8$

Putting values of $a^{2} and b^{2} in (i)$ , we have $\frac{x^{2}}{8} + \frac{y^{2}}{{(4)}^{2}} = 1 \Rightarrow \frac{x^{2}}{8} + \frac{y^{2}}{16} = 1$

Thus, the equation of the required ellipse is $\frac{x^{2}}{8} + \frac{y^{2}}{16} = 1$ .

EXERCISE

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