JEE Main 2024 (Online) 31st January Morning ShiftQuestion 1
Let the foci and length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, $a>b$ be $(\pm5,0)$ and $\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola $\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2}b^{2}}=1$ equals $\_\_\_\_\_\_$ .
Hyperbola
Answer$51$
Solution
Ellipse: $ae=5$, $\dfrac{2b^2}{a}=\sqrt{50}\Rightarrow b^2=\dfrac{5\sqrt2}{2}a$, and $a^2-b^2=25$. $a^2-\dfrac{5\sqrt2}{2}a-25=0\Rightarrow a=5\sqrt2$, so $a^2=50$, $b^2=25$. Hyperbola $\dfrac{x^2}{25}-\dfrac{y^2}{1250}=1$: $e^2=1+\dfrac{1250}{25}=51$.