Respuesta :
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) :[tex]x^2+y^2+4z^2=36[/tex]
Parametric equations:
x=-4t-1
y=2t+1
z=8t+3
Finding the gradient of function
[tex]\nabla f(x,y,z)=<\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},\frac{\partial{f}}{\partial{z}}>\\\nabla f(x,y,z)=<2x,-2y,8x>[/tex]
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
[tex]\nablaf(x,y,z)=(2x,2y,8z)=\lambda(-4,2,8)[/tex]
So, [tex]2x=-4\lambda[/tex]
[tex]\Rightarrow x=-2\lambda[/tex]
[tex]2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda[/tex]
Substitute the value of x , y and z in the ellipsoid equation
[tex](2\lambda)^2+(\lambda)^2+4(\lambda)^2=36\\9(\lambda)^2=36\\\lambda^2=4\\\lambda=\pm 2[/tex]
With [tex]\lambda = 2[/tex]
x=-2(2)=-4
y=2
z=2
With[tex]\lambda =- 2[/tex]
x=-2(-2)=4
y=-2
z=-2
Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)
