I can't solve the following question : Write the coefficient D of tangent plane $ x+By+Cz=D $ to the surface defined by equation $ 2x^2+3y^2+z^2=13/6 $ parallel to the plane $ x+y+z=1 $. Assume that the intersection of the plain and the surface happens in the positive octant.
So I calculate the gradient of the surface's equation, let's write $ \nabla (f) = 4x +6y + 2z $.
It needs to be a multiple of plane equation $ x+y+z=1 $ so we get the following equation system:
$ 4x= \lambda ; 6y = \lambda ; $ and $2z= \lambda $
I solve this equation and I get $(\frac{\lambda}{4};\frac{\lambda}{6};\frac{\lambda}{2}) $ but then what I am supposed to do ?
I read that we need to plug it into the equation but why ?
thanks for any help.
as helper told me, answer is 1.99
HINT: As the point is on your curve,
$2(\frac{\lambda}{4})^2 +3 (\frac{\lambda}{6})^2 + (\frac{\lambda}{2})^2 = \frac{13}{6}$
Find $\lambda$.
The equation of the tangent plane will be given by
$(x - \frac{\lambda}{4}) + (y - \frac{\lambda}{6}) + (z - \frac{\lambda}{2}) = 0$
You can resolve it to find the value of $D$.