Question

Find the equation of tangent of curve y²=16x which passes through point (2,8).?

Answer 1

Calculate the slope of the curve.

y²=16x

2y(dy/dx) =16

ydy/dx =8

dy/dx =8/y

Slope at (2,8)

m =dy/dx at (2,8)

    =8/8

     =1

Calculate the tangent equation.

(y-y_1) = m(x-x_1)

(y-8)=1(x-2)

y-8=x-2

x-y+6=0