Hyperbola
Ellipse Foci as Hyperbola Foci — Latus Rectum
nta_pyq_2026_jan
Grade 11
Question:
Let the foci of a hyperbola coincide with the foci of the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:
16
$24\sqrt{5}$
12
$\dfrac{96}{\sqrt{5}}$
Step-by-Step Solution
Key Concept: Ellipse: $a=6$, $b=4$, $c=\sqrt{36-16}=2\sqrt{5}$. Foci $(\pm2\sqrt{5},0)$. Hyperbola eccentricity $e=5$: $A=\tfrac{c}{e}=\tfrac{2\sqrt{5}}{5}=\tfrac{2}{\sqrt{5}}$.
$A=\tfrac{2}{\sqrt{5}}$, $B^2=\tfrac{96}{5}$. Latus rectum $=\tfrac{96}{\sqrt{5}}$.
Correct Answer: 4