10-2. Parabola, Ellipse, Hyperbola
medium

माना दीर्घवृत्त $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b$, की उत्केन्द्रता $\frac{1}{4}$ है। यदि यह दीर्घवृत्त बिन्दु $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$ से गुजरता है तो $a ^2+ b ^2$ बराबर होगा।

A

$31$

B

$29$

C

$32$

D

$34$

(JEE MAIN-2022)

Solution

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 a > b$

$e^{2}=1-\frac{b^{2}}{a^{2}}$

$\frac{1}{16}=1-\frac{b^{2}}{a^{2}}$

$\frac{b^{2}}{a^{2}}=1-\frac{1}{16}=\frac{15}{16} \Rightarrow b^{2}=\frac{15}{16} a^{2}$

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

$\frac{16 \times \frac{2}{5}}{a^{2}}+\frac{9}{b^{2}}=1$

$\frac{32}{5 a^{2}}+\frac{9}{b^{2}}=1$

$\frac{32}{5 a^{2}}+\frac{9}{\frac{15}{16} a^{2}}=1$

$\frac{80}{b^{2}}=1$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.