Question

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.

Answer 1

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

$\implies a^{2}-2b^{2}+2c^{2}-8=0$

Now $(f_{x},f_{y},f_{z})=(2x,-4y,4z)$ and so at point P it is: $(2a,-4b,4c)$ .

So, Equation of the tangent plane is $2a(x-a)-4b(y-b)+4c(z-c)=0$

$\implies ax−2by+2cz=8$

The system $2x+3y+2z = 8, x-y+2z=5$  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 $a^{2}-2b^{2}+2c^{2}-8=0$

$\implies a^2-\frac{\left(40-23a\right)^2}{8}+\frac{\left(8-3a\right)^2}{2}-8\:=0$,

$\implies -485a^2+1648a-1408=0$

$\implies a=\frac{824}{485}-i\frac{8\sqrt{61}}{485},\:a=\frac{824}{485}+i\frac{8\sqrt{61}}{485}$

we do not find a real solution for a.

Answer: The tangent plane does not exist.