Question

Find the equations of the tangent and normal to the parabola $y^2=8x$ at $(2,4)$.

Answer 1

$y^2=8x$

$2y\frac{dy}{dx}=8\Rightarrow y\frac{dy}{dx}=4$

$\Rightarrow \frac{dy}{dx}=\frac{4}{y}$

slope at (2,4) is $\frac{dy}{dx}=\frac{4}{4}=1$

Equation of normal $y-y_1=m(x-x_1)$

$y-4=1(x-2)$

$y-4=x-2$ (or)

$x-y+2=0$

Equation of normal $y-y_1=\frac{-1}{m}(x-4)$

$y-4=\frac{-1}{1}(x-2)$

$y-4=-x+2$

(or) $x+y-6=0$