Hyperbola
Point on Hyperbola — Area of Triangle PSS'
nta_pyq_2026_jan
Grade 11
Question:
Let $P(10,2\sqrt{15})$ be a point on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, whose foci are $S$ and $S'$. If the length of its latus rectum is 8, then the square of the area of $\triangle PSS'$ is equal to:
Step-by-Step Solution
Key Concept: Latus rectum $=2b^2/a=8\Rightarrow b^2=4a$. $P$ on hyperbola: $100/a^2-60/b^2=1$. Substituting: $100/a^2-15/a=1\Rightarrow a^2+15a-100=0\Rightarrow a=5$, $b^2=20$.
Area $=30\sqrt{3}$. Area$^2=2700$.
Correct Answer: 4