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10-2. Parabola, Ellipse, Hyperbola
easy
An ellipse passes through the point $(-3, 1)$ and its eccentricity is $\sqrt {\frac{2}{5}} $. The equation of the ellipse is
A
$3{x^2} + 5{y^2} = 32$
B
$3{x^2} + 5{y^2} = 25$
C
$3{x^2} + {y^2} = 4$
D
$3{x^2} + {y^2} = 9$
Solution
(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 $3{x^2} + 5{y^2} = 32$.
Standard 11
Mathematics