How do you calculate the gradient of tangent to xy=4 and x^2=16y?

Question

1427 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-11T08:32:30

At a point #(x_1,y_1)#, the gradient is #-4/x_1^2#, when it is on curve #xy=4# and #x_1/8#, when it is on #x^2=16y#.

Gradient of a curve is given by the slope of curve at that point, whose value is value of the derivative at that point.

(1) Here #xy=4# i.e. #y=4/x# and #(dy)/(dx)=-4/x^2#.

Now at point #(x_1,y_1)# on the curve (observe #x_1y_1=4#),

the value of derivative of gradient of curve is #-4/x_1^2#

(1) Here #x^2=16y# or #y=x^2/16# and #(dy)/(dx)=x/8#.

Now at point #(x_1,y_1)# on the curve (observe #x_1^2=16y_1#),

the value of derivative of gradient of curve is #x_1/8#