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उस दीर्घवृत्त, जिसके अक्ष निर्देशांक अक्ष है, जो बिन्दु $(-3,1)$ से होकर जाता है तथा जिसकी उत्केन्द्रता $\sqrt{\frac{2}{5}}$ है, का समीकरण है:
$5{x^2} + 3{y^2} - 48 = 0$
$\;3{x^2} + 5{y^2} - 15 = 0$
$\;5{x^2} + 3{y^2} - 32 = 0$
$\;3{x^2} + 5{y^2} - 32 = 0$
Solution
Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
As it passses through $(-3,1)$ we get,
$\frac{9}{a^{2}}+\frac{1}{b^{2}}=2$
$\Rightarrow 9 b^{2}+a^{2}=a^{2} b^{2}$
$=9 a^{2}\left(1-e^{2}\right)+a^{2}=a^{2} \cdot a^{2}\left(1-e^{2}\right)$
$=9 a^{2}\left(1-\frac{2}{5}\right)+a^{2}=a^{4}\left(1-\frac{2}{5}\right)$
$\Rightarrow a^{2}=\frac{32}{5}$
Now $b^{2}=a^{2}\left(1-e^{2}\right)$
$\Rightarrow b^{2}=\frac{32}{3}\left(1-\frac{2}{5}\right)$
$=\frac{32}{5}$
Hence the equation of required ellipse is
$\frac{x^{2}}{\frac{32}{3}}+\frac{y^{2}}{\frac{32}{5}}=1$
(or) $3 x^{2}+5 y^{2}=32$
