Question

SHOW THAT THE PLANE AX+BY+CZ+D=0 TOUCHES THE 3X^2-6Y^2+9Z^2+17=0 FIND THE POINT OF CONTACT

Answer 1

Given hyperboloid

$3x^2-6y^2+9z^2=-17\\dividing\ by\ 9$

$\frac{x^2}{3} - \frac{2y^2}{3} +\frac{ z^2 }{1} = \frac{- 17}{9}$

Tangential plane at any point $(a,b,c)\ is\ given\ by\\$

$\frac{ax}{3}-\frac{2by}{3}+cz=\frac{-17}{9}$

$\frac{-3ax}{17}+\frac{6by}{17}-\frac{9cz}{17}=1\ \ \ \ .......(1)$

Given plane is

$Ax+By+Cz=-D$

Or,

$\frac{-Ax}{D}-\frac{By}{D}-\frac{Cz}{D}=1\ \ \ \ \ .......(2)$

Comparing coefficient of $(1) \ and\ (2)$

We get

$a=\frac{17A}{3D}$

$b=-\frac{17B}{6D}\\c=\frac{17C}{9D}$

So,

Point of contact = $(a,b,c)=(\frac{17A}{3D},-\frac{17B}{6D},\frac{17C}{9D})$