Question

The length of the straight line x − 3y = 1 intercepted by the hyperbola x² − 4y² = 1 is

(a) $\frac{6}{\sqrt{5}}$

(b) $3\sqrt{\frac{2}{5}}$

(c) $6\sqrt{\frac{2}{5}}$

(d) none of these

Answer 1

(c) $6\sqrt{\frac{2}{5}}$

The point of intersection of $x-3y=1$and the hyperbola $x^2-4y^2=1$ is calculated in the following way:
$$\begin{gathered} {\left(1+3y\right)}^2-4y^2=1 \\ \Rightarrow 1+6y+9y^2-4y^2=1 \\ \Rightarrow 5y^2+6y=0 \\ \Rightarrow y=0\mathrm{or}y=-\frac{6}{5} \end{gathered}$$
If $y=0$, then $x=1$.
If $y=-\frac{6}{5}$, then $x=1+3\times \left(-\frac{6}{5}\right)=-\frac{13}{5}$.

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

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