Tangent plane to a surface parallel to a plane

Question

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

score 2 · Accepted Answer · answered Nov 22 '20 at 17:09

2

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 $D$.

answered Nov 22 '20 at 17:09

Math Lover

51,819

yes I mistakenly write this error. So good answer is D=2 ? Because we get $\sqrt 52/11 $ *(1/4+1/6+1/2) right ? – zzz247 Nov 22 '20 at 17:32
1

$11$ is inside the square root. You should get $\frac {\sqrt{143}} {6}$ – Math Lover Nov 22 '20 at 17:37
1

You may want to write as is and not make it $2$. – Math Lover Nov 22 '20 at 17:38
1

Btw pls also note, there are two such planes parallel to $x+y+z = 1$ on two sides of ellipsoid and you answer both. So pls retain $\pm$ with the value of $D$. – Math Lover Nov 22 '20 at 17:44

Tangent plane to a surface parallel to a plane

1 Answers1