Question

A line $3 x+y=8$ touches a hyperbola $H=0$ at $P(1,5)$ meets its asymptotes at $A$ and $B$. If $A B=2 \sqrt{10}, C(1,1)$ be the centre of hyperbola, $e$ and $l$ are eccentricity and latus rectum of hyperbola then

• A. $e=\frac{\sqrt{7}}{2}$
• B. $e=\frac{\sqrt{5}}{2}$
• C. $l=\sqrt{2}$
• D. $l=2 \sqrt{2}$

Answer 1

Answer: (B), (C)

By properties $P A=P B=\sqrt{10}$
$\frac{x 1}{1 / \sqrt{10}}-\frac{y-5}{3 / \sqrt{10}}- \pm \sqrt{10} ; A-(0,8) ; B-(2,2)$
Now, $2 \tan ^{-1} \frac{b}{a}-\tan ^{-1} \frac{4}{3}-\frac{\frac{2 b}{a}}{1-\frac{b^2}{a^2}}-\frac{4}{3} ; \frac{b}{a}-\frac{1}{2}$
$\frac{b^2}{a^2}=e^2-1 \Rightarrow e=\frac{\sqrt{5}}{2}$
Also, area of $\triangle C A B=a b: a b=\frac{1}{2}\left|\left(\begin{array}{ll}8 & 2\end{array}\right)\right| 0\left(\begin{array}{ll}2 & 1\end{array}\right)\left|2\left(\begin{array}{ll}1 & 8\end{array}\right)\right|=4$
$b^2-2, a-2 b-2 \sqrt{2} \text {, length of latus rectum }-\frac{2 b^2}{a}-\frac{4}{2 \sqrt{2}}-\sqrt{2}$