Question

The normal to the curve $2x^2+y^2=12$ at the point $(2,2)$ cuts the curve again at

• A. $(-\frac{22}{9},-\frac{2}{9})$
• B. $(\frac{22}{9},\frac{2}{9})$
• C. $(-2,-2)$
• D. None of these

Answer 1

The correct option is A $(-\frac{22}{9},-\frac{2}{9})$

$2x^2+y^2=12\Rightarrow \frac{dy}{dx}=-\frac{2x}{y}$
Slope of normal at point $(2,2)$ is $\frac{1}{2}$
Equation of normal is

$y-2=\frac{1}{2}(x-2)$
$x-2y+2=0$
$x=2y-2$
Substituting $x$ in curve equation
$2(2y-2{)}^2+y^2=12$
$9y^2-16y-4=0$
we will get $y=2,-\frac{2}{9}$
$\Rightarrow x=2,-\frac{22}{9}$
Hence the normal meets the curve again at point $(-\frac{22}{9},-\frac{2}{9})$