The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $x= \pm \frac{4}{\sqrt{13}}$, respectively. Let the line $y-\sqrt{3} x+\sqrt{3}=0$ touch this hyperbola at $\left(x_0, y_0\right)$. If $\mathrm{m}$ is the product of the focal distances of the point…

Question

The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $x= \pm \frac{4}{\sqrt{13}}$, respectively. Let the line $y-\sqrt{3} x+\sqrt{3}=0$ touch this hyperbola at $\left(x_0, y_0\right)$. If $\mathrm{m}$ is the product of the focal distances of the point $\left(x_0, y_0\right)$, then $4 \mathrm{e}^2+\mathrm{m}$ is equal to ________

Aicharya · Accepted Answer

Given $\frac{2 \mathrm{~b}^2}{\mathrm{a}}=9$ and $\frac{\mathrm{a}}{\mathrm{e}}= \pm \frac{4}{\sqrt{3}}$ equation of tangent $y-\sqrt{3} x+\sqrt{3}=0$ by equation of tangent Let slope $=\mathrm{S}=\sqrt{3}$ Constant $=-\sqrt{3}$ By condition of tangency $\begin{aligned} & \Rightarrow 6=6 \mathrm{a}^2-9 \mathrm{a} \\ & \Rightarrow \mathrm{a}=2, \mathrm{~b}^2=9 \end{aligned}$
Equation of Hyperbola is $\frac{x^2}{4}-\frac{y^2}{9}=1$ and for tangent Point of contact is $(4,3 \sqrt{3})=\left(\mathrm{x}_0, \mathrm{y}_0\right)$
Now e $=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$ Again product of focal distances $\begin{aligned} & \mathrm{m}=\left(\mathrm{x}_0 \mathrm{e}+\mathrm{a}\right)\left(\mathrm{x}_0 \mathrm{e}-\mathrm{a}\right) \\ & \mathrm{m}+4 \mathrm{e}^2=20 \mathrm{e}^2-\mathrm{a}^2 \\ & =20 \times \frac{13}{4}-4=61 \\ & \end{aligned}$
(There is a printing mistake in the equation of directrix $x= \pm \frac{4}{\sqrt{3}}$. Corrected equation is $x= \pm \frac{4}{\sqrt{13}}$ for directrix, as eccentricity must be greater than one, so question must be bonus)

The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $x= \pm…

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