Question

Find the equation of the tangent plane to the conicoid x²+y²=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

1. In the case of k = 0 we have the Oz axis as the solution.

 

2. Represent the conicoid in an implicit form:

$F(x,y,z)=x²+y²-kz=0.$

 

3. The normal vector to the surface at the point M(k,k,2k) is determined by the partial derivatives:

$\bold{n}(M) =(A,B,C)=(\frac{∂F}{∂x},\frac{∂F}{∂y},\frac{∂F}{∂z}) =(2x,2y,-k)=(2k,2k,-k).$

 

4. Thus, the equation of the tangent plane to the conicoid at the point M will by:

$A(x-x_0)+B(x-x_0)+C(z-z_0)=0;$

$2k(x-k)+2k(y-k)-k(z-2k)=0;$

$2(x-k)+2(y-k)-(z-2k)=0;$

$2x+2y-z-2k=0.$

 

5. Now, represent the plane and the conicoid geometrically and their sections with the plane x-y=0 for different values of k: