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Q. The equation of normal to the curve $y=\frac{2}{x^{2}}$ at the point on the curve where $x=1$, is

KEAMKEAM 2020

Solution:

$y=\frac{2}{x^{2}} x=1$
$\Rightarrow y=2$
$\frac{d y}{d x}=\frac{-4}{x^{3}}, m,=-4$
$\Rightarrow m_{2}=\frac{1}{4}$
$y-2=\frac{1}{4}(x-1)$
$4 y-8=x-1$
$x-4 y+7=0$