Hyperbola
Hyperbola-Ellipse Eccentricity Relation — Chord Length
nta_pyq_2024_jan
Grade 11
Question:
Let $e_1$ be the eccentricity of the hyperbola $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$ and $e_2$ be the eccentricity of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a>b$, which passes through the foci of the hyperbola. If $e_1e_2=1$, then the length of the chord of the ellipse parallel to the $x$-axis and passing through $(0,2)$ is:
$4\sqrt{5}$
$\dfrac{8\sqrt{5}}{3}$
$\dfrac{10\sqrt{5}}{3}$
$3\sqrt{5}$
Step-by-Step Solution
Key Concept: Hyperbola: $e_1=5/4$. $e_2=1/e_1=4/5$. Foci of hyperbola at $(\pm5,0)$: ellipse passes through $a=5$. Use $e_2=4/5$: $b^2=a^2(1-e_2^2)=25(1-16/25)=9$. Ellipse: $x^2/25+y^2/9=1$. Chord at $y=2$.
$e_1=5/4$, $e_2=4/5$. Ellipse: $\frac{x^2}{25}+\frac{y^2}{9}=1$. At $y=2$: $x=\pm\frac{5\sqrt5}{3}$. Chord$=\frac{10\sqrt5}{3}$.
Correct Answer: 3