Question

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

• A. $\sqrt{10}$
• B. $\frac{6}{5}$
• C. $\frac{1}{\sqrt{10}}$
• D. $\frac{6}{5}\sqrt{10}$

Answer 1

The correct option is D

$\frac{6}{5}\sqrt{10}$

Step 1 : Find the co-ordinates of point of intersection of hyperbola and straight line

Given, straight line and hyperbola are,

$x-3y=1...\left(i\right)$

$x^2-4y^2=1...\left(ii\right)$

From $\left(i\right)$ , $x=3y+1$

Substituting this value of $x$ in $\left(ii\right)$ we get

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

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

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

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

Resubstitute $y$ in $\left(i\right)$

When $y=0,x=1$

When $y=\frac{-6}{5},x=\frac{-13}{5}$

Hence the co-ordinates of the point of intersection are $\left(1,0\right)$ and $\left(\frac{-13}{5},\frac{-6}{5}\right)$

Step 2: Find the length of the intercept

Applying distance formula

Length of the intercept $=\sqrt{{\left(1+\frac{13}{5}\right)}^2+{\left(0+\frac{6}{5}\right)}^2}$

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

$=\sqrt{\frac{36\times 10}{25}}$

$\therefore$Length of the intercept $=\frac{6}{5}\sqrt{10}$

Hence option $\left(D\right)$ is the correct answer.