Question

A line $3x+y=8$ touches a hyperbola $H=0$ at $P(1,5)$ meets its asymptotes at $A$ and $B$. If $AB=2\sqrt{10},C(1,1)$ 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 $PA=PB=\sqrt{10}$
$\frac{x1}{1/\sqrt{10}}-\frac{y-5}{3/\sqrt{10}}-\pm \sqrt{10};A-(0,8);B-(2,2)$
Now, $2{\tan }^{-1}\frac{b}{a}-{\tan }^{-1}\frac{4}{3}-\frac{\frac{2b}{a}}{1-\frac{b^2}{a^2}}-\frac{4}{3};\frac{b}{a}-\frac{1}{2}$
$\frac{b^2}{a^2}=e^2-1\Rightarrow e=\frac{\sqrt{5}}{2}$
Also, area of $△CAB=ab:ab=\frac{1}{2}\left|\left(\begin{array}{ll}8& 2\end{array}\right)\right|0\left(\begin{array}{ll}2& 1\end{array}\right)\left|2\left(\begin{array}{ll}1& 8\end{array}\right)\right|=4$
$b^2-2,a-2b-2\sqrt{2}\text{, length of latus rectum }-\frac{2b^2}{a}-\frac{4}{2\sqrt{2}}-\sqrt{2}$