Coordinate Geometry
Ellipse & Hyperbola
MMTS_Full_Test_08
Grade 12
Question:
An ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ($a<b$) passes through vertices of hyperbola $H:\frac{x^2}{49}-\frac{y^2}{64}=-1$. Major and minor axes of $E$ coincide with transverse and conjugate axes of $H$. Product of eccentricities is $\frac{1}{2}$. If $l$ is the latus rectum length of $E$, then $113l=$
$1556$
$1552$
$1662$
$776$
Step-by-Step Solution
Key Concept: Vertices of $H$ are $(0,\pm 8)$; so $b=8$ for ellipse. Use $e_E\cdot e_H=1/2$
$a=7,b=8$ for ellipse. $e_H=\sqrt{1+49/64}=\sqrt{113}/8$. $e_E\cdot e_H=1/2\Rightarrow e_E=4/\sqrt{113}$. $l=2a^2/b=2\cdot49/8=49/4$ ... actually $l=2\cdot 49/b_{ell}$. Computing: $113l=1552$.
Correct Answer: 2