Question

Find the equation of normal to the curve at the point on it. 

y = x² + 2e^x + 2 at (0, 4)

Answer 1

The slope of the normal at (0, 4)

= `(-1)/(((dy)/(dx))_(at(0, 4))) = - 1/2`

∴ the equation of the normal at (0, 4) is

`y - 4 = -1/2 (x - 0)`

∴ 2y − 8 = −x

∴ x + 2y − 8 = 0

Hence, the equations of tangent and normal are 2x − y + 4 = 0 and x + 2y − 8 = 0 respectively.