find the equation of the normal of the curve y^2-x^2=5 at poin (2,3).?

Answer 1

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 2

By implicit differentiation
2y*y´-2x= 0 so at point( 2,3)
6y´-4=0 so y´= 2/3 and the normal is

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

Answer 3

2y.dy/dx - 2x = 0
dy/dx = x/y
Gradient of tangent at (2,3) = 2 / 3
Gradient of normal = - 3 / 2
Equation of normal is:-
y - 3 = - (3/2).(x - 2)
y = - (3/2).x + 6