Question

Find a point on the ellipsoid x2+y2+4z2=36x2+y2+4z2=36 where the tangent plane is perpendicular to the line with parametric equations

Answer 1

Answer:

A point on the ellipsoid is (-4,2,2) or (4,-2,-2)

Step-by-step explanation:

Given equation of ellipsoid f(x,y,z) :$x^2+y^2+4z^2=36$

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

$\nabla f(x,y,z)=\\\nabla f(x,y,z)=$

So, The directions vectors=(-4,2,8)

Now the line is perpendicular to plane when direction vector is parallel to the normal vector of line

$\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)$

So, $2x=-4\lambda$

$\Rightarrow x=-2\lambda$

$2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda$

Substitute the value of x , y and z in the ellipsoid equation

$(2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2$

With $\lambda = 2$

x=-2(2)=-4

y=2

z=2

With$\lambda =- 2$

x=-2(-2)=4

y=-2

z=-2

Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)