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](https://tex.z-dn.net/?f=x%5E2%2By%5E2%2B4z%5E2%3D36)
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)=](https://tex.z-dn.net/?f=%5Cnabla%20f%28x%2Cy%2Cz%29%3D%3C%5Cfrac%7B%5Cpartial%7Bf%7D%7D%7B%5Cpartial%7Bx%7D%7D%2C%5Cfrac%7B%5Cpartial%7Bf%7D%7D%7B%5Cpartial%7By%7D%7D%2C%5Cfrac%7B%5Cpartial%7Bf%7D%7D%7B%5Cpartial%7Bz%7D%7D%3E%5C%5C%5Cnabla%20f%28x%2Cy%2Cz%29%3D%3C2x%2C-2y%2C8x%3E)
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)](https://tex.z-dn.net/?f=%5Cnablaf%28x%2Cy%2Cz%29%3D%282x%2C2y%2C8z%29%3D%5Clambda%28-4%2C2%2C8%29)
So, ![2x=-4\lambda](https://tex.z-dn.net/?f=2x%3D-4%5Clambda)
![\Rightarrow x=-2\lambda](https://tex.z-dn.net/?f=%5CRightarrow%20x%3D-2%5Clambda)
![2y=2\lambda\\\Rightarrow y=\lambda\\8z=8\lambda\\\Rightarrow z=\lambda](https://tex.z-dn.net/?f=2y%3D2%5Clambda%5C%5C%5CRightarrow%20y%3D%5Clambda%5C%5C8z%3D8%5Clambda%5C%5C%5CRightarrow%20z%3D%5Clambda)
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](https://tex.z-dn.net/?f=%282%5Clambda%29%5E2%2B%28%5Clambda%29%5E2%2B4%28%5Clambda%29%5E2%3D36%5C%5C9%28%5Clambda%29%5E2%3D36%5C%5C%5Clambda%5E2%3D4%5C%5C%5Clambda%3D%5Cpm%202)
With ![\lambda = 2](https://tex.z-dn.net/?f=%5Clambda%20%3D%202)
x=-2(2)=-4
y=2
z=2
With![\lambda =- 2](https://tex.z-dn.net/?f=%5Clambda%20%3D-%202)
x=-2(-2)=4
y=-2
z=-2
Hence a point on the ellipsoid is (-4,2,2) or (4,-2,-2)