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)

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.