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

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since it passes through $(-3, 1)$,we have $\frac{9}{a^2} + \frac{1}{b^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since it passes through $(-3, 1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$.Multiplying by $a^2$,we get $9 + \frac{a^2}{b^2} = a^2$.....$(i)$Given the eccentricity $e = \sqrt{\frac{2}{5}}$,we know $e^2 = 1 - \frac{b^2}{a^2}$.So,$\frac{2}{5} = 1 - \frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2} = \frac{3}{5}$,which means $\frac{a^2}{b^2} = \frac{5}{3}$.....$(ii)$Substituting $(ii)$ into $(i)$,we get $9 + \frac{5}{3} = a^2 \Rightarrow a^2 = \frac{27 + 5}{3} = \frac{32}{3}$.From $(ii)$,$b^2 = \frac{3}{5} a^2 = \frac{3}{5} 	imes \frac{32}{3} = \frac{32}{5}$.Substituting $a^2$ and $b^2$ into the standard equation: $\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$.This simplifies to $\frac{3x^2}{32} + \frac{5y^2}{32} = 1$,or $3x^2 + 5y^2 = 32$.

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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