Question

Find the equation of the tangents to the hyperbola $x^2-4y^2=4$ which are :
i) Parallel ii) Perpendicular to the line $x+2y=0$.

Answer 1

Given hyperbola can be written as, $\frac{x^2}{4}-\frac{y^2}{1}=1$

$\Rightarrow a^2=4,b^2=1$

And slope of line $x+2y=0$ is $-\frac{1}{2}$

(i)Slope of tangent $=m=-\frac{1}{2}$

Now using condition of tangency, $c^2=a^2{m}^2-b^2$

$\Rightarrow c^2=1-1=0\Rightarrow c=0$

So equation of tangents is $y=mx+c=-\frac{1}{2}x+0\Rightarrow x+2y=0$

(ii) Slope of tangent $,m=2$

Now using condition of tangency, $c^2=a^2{m}^2-b^2$

$\Rightarrow c^2=16-1=15\Rightarrow c=\pm \sqrt{15}$

So equation of tangents is $y=mx+c$

$\Rightarrow y=2x\pm \sqrt{15}$