Question

Show that the line $x-y+4=0$ is a tangent to the ellipse $x^2+3y^2=12$. Find the point of contact

Answer 1

$x^2+3y^2=12\Rightarrow \frac{x^2}{12}+\frac{y^2}{4}=1(÷$ by $12)$
Here $a^2=12,b^2=4$
$\therefore$ the condition for $y=mx+c$ to be a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $c^2=a^2{m}^2+b^2$
$x-y+4=0\Rightarrow y=x+4$
Here, $m=1,c=4\Rightarrow c^2=4^2=16$
$a^2{m}^2÷b^2=12(1{)}^2+4=12+4=16$
$\therefore c^2=a^2+b^2=16$
Hence, $x-y+4=0$ is a tangent to the ellipse
$\therefore$ The point of contact $=(\frac{a^{2}m}{c},\frac{b^2}{c})=(\frac{-12\left(1\right)}{4},\frac{4}{4})$ $=(-3,1)$