Question

find the equation of the normal to the curve y=3√(x+1) at the point where x=

Answer 1

x=8

Answer 2

Normal : The normal at the point P of a curve C is the line through P perpendicular to the tangent at P.

This curve is a hyperbola centered at the origin.

Solving for y:

y = SQRT(5 + x^2)
in the first quadrant
(since (2, 3) is located in quad I).

The derivative is y' = 1/2(5 + x^2)^(-1/2)(2x)

The slope of the tangent at (2, 3) is...

y'(2) = 1/2(5 + 2^2)^(-1/2)(2)(2)

=(1/2)(9)^(-1/2)(4)
=2/(9^1/2)
= 2/3

The slope of the normal is perpendicular to the tangent, so the slope of the normal is -3/2

Using the point (2, 3), the equation for the normal is

(y - 3) / (x - 2) = -3/2

Answer 3

nooo

Answer 4

where did u get 5+x^2

Answer 5

similarly u can solve by placing 3sqrootx+1 instead of 5+x^2

Answer 6

YEAH

Answer 7

method to solve has sshown up by abayomi then u can proceed further