Let the sum of the focal distances of the point $\mathrm{P}(4,3)$ on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ be $8 \sqrt{\frac{5}{3}}$. If for $H$, the length of the latus rectum is $l$ and the product of the focal distances of the point P…

Question

Let the sum of the focal distances of the point $\mathrm{P}(4,3)$ on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ be $8 \sqrt{\frac{5}{3}}$. If for $H$, the length of the latus rectum is $l$ and the product of the focal distances of the point P is m , then $9 l^2+6 \mathrm{~m}$ is equal to :-

Aicharya · Accepted Answer

$\begin{aligned} & e x+a+e x-a=8 \sqrt{\frac{5}{3}} \\ & 2 e x=8 \sqrt{\frac{5}{3}} \\ & 2 e \times 4=8 \sqrt{\frac{5}{3}} \\ & e=\sqrt{\frac{5}{3}} \\ & b^2=a^2\left(\left(\frac{\sqrt{5}}{3}\right)^2-1\right) \\ & b^2=\frac{2}{3} a^2\end{aligned}$
$\frac{16}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1$
and $b^2=\frac{2}{3} a^2$
$\Rightarrow \mathrm{a}^2=\frac{5}{2} \quad \mathrm{~b}^2=\frac{5}{3}$
Now,
$\begin{aligned}
& \ell=\frac{2 \mathrm{~b}^2}{\mathrm{a}} \\ & \ell^2=\frac{4 \mathrm{~b}^4}{\mathrm{a}^2} \\ & 9 \ell^2=36 \times \frac{25}{9 \times 5} \times 2 \\ & 9 \ell^2=40 \\ & \mathrm{~m}=(\mathrm{ex}+\mathrm{a})(\mathrm{ex}-\mathrm{a}) \\ & \mathrm{m}=\mathrm{e}^2 \mathrm{x}^2-\mathrm{a}^2 \\ & =\frac{5}{3} \times 16-\frac{5}{2}=\frac{145}{6}
\end{aligned}$
$\begin{aligned}
& =6 \mathrm{~m}=145 \\ & 9 \ell^2+6 \mathrm{~m} \\ & 40+145=185
\end{aligned}$
option (3)

Let the sum of the focal distances of the point $\mathrm{P}(4,3)$ on the hyperbola $\mathrm{H}:…

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