Question

Find the point on the curve x2/9-y2/16=1 at which the tangent is parallel to the y axis

Answer 1

Answer:

x^2/9-y2/16=1

differntiating both side

2x/9-2y/16y'=0

y'=16x/9y

Given that y'=1/0

y=0,x=+-3

Hence points are

(3,0);(-3,0)