Question

Find the equation of the normal to curve $y^2=4x$ at the point (1, 2).

Answer 1

The equation of the given curve is $y^2=4x$
On differentiating w.r.t. x, we get
$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}$
$\therefore$ Slope of tangent at (1,2), is $(\frac{dy}{dx}{)}_{(1,2)}=\frac{2}{2}=1$
Slope of normal at the point $(1,2)=-\frac{1}{1}=-1$
$\therefore$ Equation of the normal at (1,2) is y - 2 = - 1(x-1)
$\Rightarrow y-2=-x+1\Rightarrow x+y-3=0$