Question

The length of the straight line $x-3y=1$ intercepted by hyperbola $x^2-4y^2=1$ is

Answer 1

Let $A(x_1,y_1)$ and $B(x_2,y_2)$ are the point of intersection of line $x-3y=1$ and hyperbola

$x^2-4y^2=1$

$\because x-3y=1$

$\Rightarrow x=3y+1\dots \left(1\right)$

Putting value of $x$ in equation of hyperbola

$\Rightarrow (3y+1{)}^2-4y^2=1$

$\Rightarrow 9y^2+6y+1-4y^2=1$

$\Rightarrow 5y^2+6y=0$

$\Rightarrow y(5y+6)=0$

$\Rightarrow y=0,-\frac{6}{5}$

$\therefore y_1=0,y_2=-\frac{6}{5}$

From equation $\left(1\right)$

When $y_1=0\Rightarrow x_1=1$

And when $y_2=-\frac{6}{5}\Rightarrow x_2=-\frac{13}{5}$

$\therefore$ Point of intersection $A(1,0)$ and $B(-\frac{13}{5},-\frac{6}{5})$

Intercepted length

$AB=\sqrt{({(-\frac{13}{5}-1)}^2+{(-\frac{6}{5}-0)}^2}$

$\Rightarrow AB=\sqrt{\frac{324}{25}+\frac{36}{25}}$

$\Rightarrow AB=\sqrt{\frac{360}{25}}$

$\Rightarrow AB=\sqrt{\frac{72}{5}}$

$\Rightarrow AB=6\sqrt{\frac{2}{5}}$

Hence, option (C) is correct.