# Find the Equation of an Ellipse Whose Axes Lie Along the Coordinate Axes, Which Passes Through the Point (−3, 1) and Has Eccentricity Equal to √ 2 / 5 - Mathematics

Find the equation of an ellipse whose axes lie along the coordinate axes, which passes through the point (−3, 1) and has eccentricity equal to $\sqrt{2/5}$

#### Solution

$\text{ Let the equation of the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ...(1)$
$\text{ It passes through the point }\left( -3,1 \right).$
$\therefore\frac{9}{a^2}+\frac{1}{b^2}=1 ...( 2)$
$\text{ and } e = \sqrt{\frac{2}{5}}$
$\text{ Now }, b^2 = a^2 \left( 1 - e^2 \right)$
$\Rightarrow b^2 = a^2 \left( 1 - \frac{2}{5} \right)$
$\Rightarrow b^2 = a^2 \times \frac{3}{5}or\frac{3 a^2}{5}$
$\text{ Substituting the value of b^2 in eq. (2), we get }:$
$\frac{9}{a^2}+\frac{5}{3 a^2}=1$
$\Rightarrow \frac{27 + 5}{3 a^2} = 1$
$\Rightarrow a^2 = \frac{32}{3}$
$\Rightarrow b^2 = \frac{3 \times \frac{32}{3}}{5} or \frac{32}{5}$
$\text{ Substituting the values ofaandbin eq. (1), we get }:$
$\frac{3 x^2}{32} + \frac{5 y^2}{32} = 1$
$\Rightarrow 3 x^2 + 5 y^2 = 32$
$\text{ This is the required equation of the ellipse }.$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 15 | Page 23