JEE Main 2024 (31 Jan, Shift 1) Maths Q16: Conic Sections

Question

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2},\frac{14}{3}\sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to

Accepted Answer

Option 3: $7\sqrt{\frac{2}{5}}-\frac{8}{3}$

Answer

$7\sqrt{\frac{2}{5}}+\frac{8}{3}$

Answer

$14\sqrt{\frac{2}{5}}-\frac{4}{3}$

Answer

$7\sqrt{\frac{2}{5}}-\frac{8}{3}$

Answer

$14\sqrt{\frac{2}{5}}-\frac{16}{3}$

JEE Main 31 January 2024, Shift 1, Maths Q16