Hyperbola
Point on Hyperbola — Focal Distances Product
nta_pyq_2024_apr
Grade 11
Question:
Let $H:\dfrac{-x^2}{a^2}+\dfrac{y^2}{b^2}=1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha,6)$, $\alpha>0$ lies on $H$. If $\beta$ is the product of the focal distances of the point $(\alpha,6)$, then $\alpha^2+\beta$ is equal to
Step-by-Step Solution
Key Concept: Hyperbola: $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$. Eccentricity $e=\sqrt{1+a^2/b^2}=\sqrt{3}\Rightarrow a^2/b^2=2\Rightarrow a^2=2b^2$. LR $=2a^2/b=4\sqrt{3}\Rightarrow a^2=2b\sqrt{3}$. Solving: $b=\sqrt{3}$, $a=\sqrt{6}$.
$b=\sqrt{3}$, $a^2=6$. $\alpha^2=66$. Foci $(0,\pm3)$. $\beta=\sqrt{147}\cdot\sqrt{75}=105$. $\alpha^2+\beta=171$.
Correct Answer: 2