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

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) :

Parametric equations:

x=-4t-1

y=2t+1

z=8t+3

Finding the gradient of function

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

So,

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

With

x=-2(2)=-4

y=2

z=2

With

x=-2(-2)=4

y=-2

z=-2

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

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step-by-step explanation: