Given surface defined by $2x^2+3y^2+z^2=\frac{13}{6}$ parallel to the plane $x+y+z=1$. Find the tangent plane corresponding to it. The intersection of the plane and the surface happens in the first quadrant.
I found the normal of the surface as <4x,6y,2z> and equated it to the normal of the plane $x+y+z=1$ and got value of x in terms of a constant k and plugged it in the equation of the surface to get k and thus values x,y and z.
But here I am getting k as irrational and the calculation is becoming complex and lengthy.
Can you please let me know if my approach to the above formulation is correct or suggest a better way for the same?
Best Answer
Matching the two normals $$4x:6y:2z=1:1:1$$ or, $y=\frac23x$ and $z=2x$. Then, substitute them into $2x^2+3y^2+z^2=\frac{13}{6}$ to get $$2x^2 +3(\frac23 x)^2+(2x)^2=\frac{13}6$$ which leads to the tangent point $( \frac12\sqrt{\frac{13}{11}},\frac13\sqrt{\frac{13}{11}}, \sqrt{\frac{13}{11}})$. Then, the tangent plane is $$(1,1,1)\cdot\left(x-\frac12\sqrt{\frac{13}{11}}, y-\frac13\sqrt{\frac{13}{11}}, z-\sqrt{\frac{13}{11}}\right)=0 $$ which is $$x+y+z= \frac{\sqrt{143}}6$$