jee-main 2025 Q25

jee-main · India · session1_24jan_shift2 Conic sections Eccentricity or Asymptote Computation
Let $\mathrm{H}_{1} : \frac{x^{2}}{\mathrm{a}^{2}} - \frac{y^{2}}{\mathrm{b}^{2}} = 1$ and $\mathrm{H}_{2} : -\frac{x^{2}}{\mathrm{A}^{2}} + \frac{y^{2}}{\mathrm{B}^{2}} = 1$ be two hyperbolas having length of latus rectums $15\sqrt{2}$ and $12\sqrt{5}$ respectively. Let their eccentricities be $e_{1} = \sqrt{\frac{5}{2}}$ and $e_{2}$ respectively. If the product of the lengths of their transverse axes is $100\sqrt{10}$, then $25\mathrm{e}_{2}^{2}$ is equal to $\_\_\_\_$.
Let $\mathrm{H}_{1} : \frac{x^{2}}{\mathrm{a}^{2}} - \frac{y^{2}}{\mathrm{b}^{2}} = 1$ and $\mathrm{H}_{2} : -\frac{x^{2}}{\mathrm{A}^{2}} + \frac{y^{2}}{\mathrm{B}^{2}} = 1$ be two hyperbolas having length of latus rectums $15\sqrt{2}$ and $12\sqrt{5}$ respectively. Let their eccentricities be $e_{1} = \sqrt{\frac{5}{2}}$ and $e_{2}$ respectively. If the product of the lengths of their transverse axes is $100\sqrt{10}$, then $25\mathrm{e}_{2}^{2}$ is equal to $\_\_\_\_$.