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