An ellipse passes through the point (-3, 1) a | vedclass

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Question

(a) Let the equation of ellipse be $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

It passes through $(-3, 1)$

So, $\frac{9}{{{a^2}}} +

Vedclass · Accepted Answer

$3{x^2} + 5{y^2} = 32$

Vedclass · Answer

(a) Let the equation of ellipse be $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

It passes through $(-3, 1)$

So, $\frac{9}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$

$\Rightarrow 9 + \frac{{{a^2}}}{{{b^2}}} = {a^2}$.....$(i)$

Given eccentricity is $\sqrt {2/5} $

So, $\frac{2}{5} = 1 - \frac{{{b^2}}}{{{a^2}}} $

$\Rightarrow \frac{{{b^2}}}{{{a^2}}} = \frac{3}{5}$.....$(ii)$

From equation $(i)$ and $(ii),$ ${a^2} = \frac{{32}}{3},{b^2} = \frac{{32}}{5}$

Hence required equation of ellipse is $3{x^2} + 5{y^2} = 32$.

An ellipse passes through the point $(-3, 1)$ and its eccentricity is $\sqrt {\frac{2}{5}} $. The equation of the ellipse is

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