Coordinate Geometry
Common foci of ellipse and hyperbola; eccentricity relations
Grade Class 12
Question:
Let points $S_1$ and $S_2$ (lying on positive $x$-axis) be the foci of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and $\dfrac{x^2}{a^2}-\dfrac{y^2}{c^2}=\dfrac{1}{4}$. If a point of intersection is equidistant from $S_1$ and $S_2$ and $a:b=\sqrt{7}:\sqrt{6}$, then $a:c$ is
$\sqrt{7}:\sqrt{6}$
$\sqrt{7}:\sqrt{18}$
$\sqrt{7}:3$
$\sqrt{7}:\sqrt{2}$
Step-by-Step Solution
Key Concept: A point equidistant from $S_1$ and $S_2$ lies on the $y$-axis (perpendicular bisector of $S_1S_2$). Use $a:b=\sqrt{7}:\sqrt{6}$ to find eccentricity $e$ of ellipse, then establish the hyperbola's eccentricity $E$.
Eccentricity of ellipse $e=1/\sqrt{7}$. Solving gives $a:c=\sqrt{7}:\sqrt{18}$.
Correct Answer: 2