Hyperbola
Latus Rectum Subtending Angle at Centre
nta_pyq_2024_jan
Grade 11
Question:
Let the latus rectum of the hyperbola $\dfrac{x^2}{9}-\dfrac{y^2}{b^2}=1$ subtend an angle of $\dfrac{\pi}{3}$ at the centre of the hyperbola. If $b^2$ is equal to $\dfrac{l}{m}(1+\sqrt{n})$, where $l$ and $m$ are co-prime numbers, then $l^2+m^2+n^2$ is equal to
Step-by-Step Solution
Key Concept: The LR endpoints are $(ae, \pm b^2/a)$. The half-angle from the axis is $\tan^{-1}\left(\frac{b^2/a}{ae}\right)=\pi/6$ (since full angle is $\pi/3$). So $\tan30°=\frac{b^2}{a^2e}=\frac{1}{\sqrt3}$. Use $e^2=1+b^2/9$ to solve.
$\tan30°=b^2/(9e)=1/\sqrt3\Rightarrow b^2=3\sqrt3e$. $e=\sqrt3b^2/9$: $e^2=b^4/27$. $e^2=1+b^2/9\Rightarrow b^4/27=1+b^2/9\Rightarrow b^4-3b^2-27=0\Rightarrow b^2=\frac{3}{2}(1+\sqrt{13})$. $l=3,m=2,n=13$. $9+4+169=182$.
Correct Answer: 182