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

Answer: (D)

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

Answer 2

Given : equation of line, $x - 3y = 1$ (1)
and hyperbola $x^2 - 4y^2 = 1$ (2)
putting $x = 1 + 3y$ in equation (2), we get
$(1 + 3y)^2 - 4y^2 = 1 \; \Rightarrow \; 1+ 9y^2 + 6y - 4y^2 = 1$
$\Rightarrow \; 5y^2 + 6y = 0 \; \Rightarrow \; y(5y + 6) = 0$
$\Rightarrow$ y = 0 or y = $- \frac{6}{5}$
$\therefore$ x = 1 for y = 0 & x = 1 - $\frac{18}{5} = \frac{-13}{5}$
for y = -6/5 $\therefore$ the line (1) cuts the hyperbola (2) in at most two point.
$\therefore$ co-ordinates of points are P(1,0) & $Q \left( - \frac{13}{5} , - \frac{6}{5} \right)$
$\therefore \; PQ = \sqrt{ \left(1+ \frac{13}{5}\right)^{2} +\left(0+\frac{6}{5}\right)^{2}}$
$= \sqrt{\left(\frac{18}{5}\right)^{2} +\frac{36}{25}} = \sqrt{\frac{324 +36}{25}} = \sqrt{\frac{360}{25}}$
$\therefore$ length of straight line $PQ = \frac{6\sqrt{10}}{5}$ units.