Answer to Question #105595 in Analytic Geometry for Vikas

Question

a) Does there exist a plane targent to x
2 −2y
2 +2z
2 = 8 and which passes through
2x+3y+2z = 8, x−y+2z = 5? Justify your answer. (5)

Answer 1

Given $f(x,y,z)=x^{2}-2y^{2}+2z^{2}-8=0$ with point $P(a,b,c)$ .

We calculate $(f_{x},f_{y},f_{z})=(2x,-4y,4z)$ in this point: $(2a,-4b,4c)$ .

Equation of the tangent plane is $2a(x-a)-4b(x-b)+4c(x-c)=0$ , $ax-2by+2cz=8.$

The system $\left\{\begin{array}{ll} 2x+3y+2z = 8 \\ x-y+2z=5 \end{array}\right.$ is a line, we choose points on the line.

The points $(3,0,1)$ and $(-1,1,\frac{7}{2})$ of the given line must lie in this tangent plane,

then we have $c=\frac{8-3a}{2}$ and $b=\frac{40-23a}{4}$ .

Using the equation $a^{2}-2b^{2}+2c^{2}-8=0$ , we do not find a real solution for $a$ .

Answer: The tangent plane does not exist.