10-2. Parabola, Ellipse, Hyperbola
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ધારો કે કોઈક ઉપવલય $\frac{x^{2}}{ a ^{2}}+\frac{y^{2}}{ b ^{2}}=1, a > b$ ની ઉત્કેન્દ્રતા $\frac{1}{4}$ છે. જો આ ઉપવલય,બિંદુ $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$ માંથી પસાર થતો હોય તો,$a^{2}+b^{2}=\dots\dots\dots$

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

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