Question

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

• A. $\frac{3\sqrt{10}}{5}$
• B. $\frac{6\sqrt{10}}{5}$
• C. $\frac{5\sqrt{10}}{3}$
• D. $\frac{5\sqrt{10}}{6}$

Answer 1

The correct option is B

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

Explanation for correct option:

Step-1 : Finding the intercept points:

Given equation of straight line is,

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

And equation of hyperbola is,

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

Solving equation $\left(1\right)$ and $\left(2\right)$ we get,

$$\begin{gathered} x-3y=1 \\ \Rightarrow x=1+3y \end{gathered}$$

Substituting this in equation $\left(2\right)$ we get,

$$\begin{gathered} {\left(1+3y\right)}^2-4y^2=1 \\ \Rightarrow 1+9y^2+6y-4y^2=1 \\ \Rightarrow 5y^2+6y=0 \\ \Rightarrow y\left(5y+6\right)=0 \\ \Rightarrow y=0,-\frac{6}{5} \end{gathered}$$:

Substituting $y=0$ in equation $\left(1\right)$ we get,

$\begin{array}{rcl}x-3y& =& 1\\ x-\left(3\times 0\right)& =& 1\\ x& =& 1\end{array}$

Substituting $y=-\frac{6}{5}$ in equation $\left(1\right)$ we get,

$$\begin{gathered} x-3y=1 \\ \Rightarrow x-3\left(-\frac{6}{5}\right)=1 \\ \Rightarrow x+\left(\frac{18}{5}\right)=1 \\ \Rightarrow x=1-\frac{18}{5} \\ \Rightarrow x=\frac{5-18}{5} \\ \Rightarrow x=-\frac{13}{5} \end{gathered}$$

Hence, the two points are $\left(1,0\right)$ and $\left(-\frac{13}{5},-\frac{6}{5}\right)$

Step-2 : Finding the length of the straight line

Now the distance between these two points is,

$$\begin{gathered} \mathrm{Distance}=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ =\sqrt{{\left(-\frac{13}{5}-1\right)}^2+{\left(-\frac{6}{5}-0\right)}^2} \\ =\sqrt{\frac{324}{25}+\frac{36}{25}} \\ =\sqrt{{\left(\frac{6}{5}\right)}^2\left(3^2+1\right)} \\ =\frac{6}{5}\sqrt{10} \end{gathered}$$

Therefore the length of a straight line intercepted by given hyperbola is equal to $\frac{6}{5}\sqrt{10}$.

Hence, option (B) is the correct answer.