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 $y^2=4x$
Differentiating with respect to x,we have:
$2y\frac{dy}{dx}=4$
⇒ $\frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}$
∴ $\frac{dy}{dx}\bigg]_{(1,2)}$= $\frac{2}{2}=1$
Now, the slope of the normal at point (1,2) is -$\frac{-1}{\frac{dy}{dx}\bigg]_{(1,2)}}$=-$-\frac{1}{1}=-1$
∴Equation of the normal at (1, 2) is y − 2 = −1(x − 1).
∴ y−2 =−x+1
∴ x+y−3=0