Question

Find an equation of the tangent plane to the given surface at the specified point. Z = 4(x - 1)2 + 4(y + 3)2 + 4, (2, -2, 12)

Answer 1

Hii, given x = a sec @ and y = b tan @
dx = a sec @ tan @ d@
And dy = b sec^2 @ d@
dy/dx is got by dividing second by the first
==> dy/dx = b sec @ / a tan @. [This is the slope]
Point x = a sec @ and y = b tan @
Hence equation of tangent at the above point
y - b tan @ = {b sec @ / a tan @} ( x - a sec @)
==> a tan @ * y - a b tan^2 @ = b x sec @ - a b sec^2 @
Rearranging ab ( sec^2 @ - tan^2 @) = x b sec @ - y a tan @
We know sec^2 @ - tan^2 @ = 1
Hence equation of tangent
x b sec @ - y a tan @ = a b [PROVED]