Question

Find the equation of the tangents to the ellipse $2x^2+y^2=8$ which are
(i) parallel to $x-2y-4=0$
(ii) perpendicular to $x+y+2=0$

Answer 1

Equaton of ellipse-

$2x^2+y^2=8$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{8}=1$

Here,

$a^2=4$

$b^2=8$

Given equation of tangent-

$x-2y-4=0$

$y=\frac{x}{2}-\frac{4}{2}$

Now,

• (i) Equation of tangent parallel to $y=\frac{x}{2}-\frac{4}{2}$ is-
  

$y=\frac{x}{2}+\frac{k}{2}.....\left(1\right)$

Here

$m=\frac{1}{2}$

$c=\frac{k}{2}$

Equation $\left(1\right)$ is tangent to the ellipse if

$c^2=a^2{m}^2+b^2$

${\left(\frac{k}{2}\right)}^2=4\times \frac{1}{4}+8$

$\frac{k^2}{4}=9$

$\Rightarrow k=\pm \sqrt{36}=\pm 6$

Hence the equation of tangents are $y=\frac{x}{2}\pm \frac{6}{2}\Rightarrow 2y=x\pm 6$

• (ii) Equation of tangent perpendicular to $y=\frac{x}{2}-\frac{4}{2}$ is-

$y=-2x+\frac{k}{2}.....\left(2\right)$

Equation $\left(1\right)$ is tangent to the ellipse if

${\left(\frac{k}{2}\right)c}^2=a^2{m}^2+b^2$

$\frac{k^2}{4}=4\times 4+8$

$\frac{k^2}{4}=96$

$k=\pm \sqrt{96}=\pm 4\sqrt{6}$

Hence the equation of tangents are $y=-2x\pm 4\sqrt{6}$