Answer to Question #86514 in Analytic Geometry for hitendra

Question

Find the equation of the tangent plane to the conicoid x^2+y^2 = kz at the point
(k;k;2k), where k is a constant. Represent the plane geometrically. Now take
different values of k, including both positive and negative, and see how the shape of
the conicoid changes.

Answer 1

We have an equation of conicoid

$$F(x,y,z) = x^2 + y^2 - kz$$

And the point

$$M(k,k,2k)$$

Let’s find a derivative

$$F’x(x,y,z) = 2x; F'x(M) = 2k$$

$$F’y(x,y,z) = 2y; F’y(M) = 2k$$

$$F’z(x,y,z) = -k; F’x(M) = -k$$

So equation of tangent plane is:

$$F’x(M)*(x-k) + F’y(M)*(y-k) + F’z(M)*(z-2k) = 0$$

$$2k*(x-k) + 2k*(y-k)*-k*(z-2k) = 2kx + 2ky - kz -2k2 = 0$$

Answer:

$$2kx + 2ky - kz -2k^2 = 0$$

if k > 0

if k < 0