Hyperbola
Tangent to Hyperbola — Focal Distances Product
nta_pyq_2024_apr
Grade 11
Question:
The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x=\pm\dfrac{4}{\sqrt{13}}$, respectively. Let the line $y-\sqrt{3}x+\sqrt{3}=0$ touch this hyperbola at $(x_0,y_0)$. If $m$ is the product of the focal distances of the point $(x_0,y_0)$, then $4e^2+m$ is equal to
Step-by-Step Solution
Key Concept: Directrix $x=\pm a/e=\pm4/\sqrt{13}$. LR $=2b^2/a=9$. Tangent $y=\sqrt{3}x-\sqrt{3}$ has slope $\sqrt{3}$, constant $-\sqrt{3}$. Condition of tangency: $(-\sqrt{3})^2=a^2(\sqrt{3})^2-b^2\Rightarrow 3=3a^2-b^2$.
$a=2$, $b^2=9$, $e=\sqrt{13}/2$. Tangent point $(4,3\sqrt{3})$. $m=20e^2-a^2=65-4=61$. $4e^2+m=13+... $ Answer is $61$ (note: printing mistake in directrix; corrected to $x=\pm4/\sqrt{13}$ and question is bonus per official solution).
Correct Answer: 61