Hyperbola
Eccentricity of Hyperbola from Ellipse Parameters
nta_pyq_2024_jan
Grade 11
Question:
Let the foci and length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$ be $(\pm5,0)$ and $\sqrt{50}$, respectively. Then the square of the eccentricity of the hyperbola $\dfrac{x^2}{b^2}-\dfrac{y^2}{a^2b^2}=1$ equals
Step-by-Step Solution
Key Concept: From ellipse: $ae=5\Rightarrow a=5$, $e=1$... wait, foci at $(\pm5,0)$: $ae=5$. LR$=2b^2/a=\sqrt{50}$. Solve for $a,b,e$. Then find eccentricity of the given hyperbola.
Ellipse: $ae=5,2b^2/a=\sqrt{50}$. Solving: $e=1/\sqrt2,a=5\sqrt2,b^2=25$. Hyperbola $\frac{x^2}{25}-\frac{y^2}{2500}=1$: $e_H^2=1+\frac{2500}{25}=1+a^2=51$.
Correct Answer: 51