The ellipsoid 4x2+2y2+z2=16 intersects the plane
y=2 in an ellipse.
Find parametric equations for the tangent line to this ellipse at the point (1,2,2).
The equation of a line in parametric form;
x = xo + at
y = yo + bt
z = zo + ct
where (xo,yo,zo) is a point on the line and (a,b,c) is the direction vector of the line
Step 1:
Find the equation of the ellipse.
Substitute y = 2 into the ellipsoid equation:
4 x^2 + 2 (2)^2 + z^2 = 16
4x^2 + 8 + z^2 = 16 (subtract 8 from both sides)
4x^2 + z^2 = 8
Step 2:
Find the partial derivatives
Find the partial derivative of the ellipsoid equation with respect to x:
d/dx (4x^2 + 2y^2 + z^2) = d/dx (16)
8x +0 +2z dz/dx = 0
dz/dx = -8x/2z = -4x/z
Find the partial derivative of the ellipsoid equation with respect to z:
d/dz (4x^2 + 2y^2 + z^2) = d/dz (16)
8x +dx/dz+ 0 + 2z = 0
dx/dz = -2z/8x = -z/4x
Step 3:
Evaluate he partial derivatives at the point (1,2,2)
Evaluate dz/dx at (1,2,2):
Find the direction vector of the tangent line.
dz/dx l (1,2,2) = - 2/4 (1) = - 1/2
Step 4:
Find he direction vector of the tangent line.
The direction vector of he tangent line is given by v magnitude = 1,0, dz/dx (1,2,2)
v magnitude = (1,0-2)
Step 5:
Write the parametric equations of the tangent line.
Using the point (1,2,2) and the direction vector (1,0,-2), the parametric equations are the following:
- x = 1 + t
- y = 2 + 0t = 2
- z= 2- 2t
Final Answer:
The parametric equations for the tangent line to the ellipse at the point (1,2,2) are the following:
- x = 1+ t
- y = 2
- z = 2 - 2t
I hope the mathematical calculations (step by step approach) was helpful. Please let me know if you require any more assistance. If anyone in my neighborhood is interested in setting up an in-person math tutoring session. I look forward to hearing from them. Have an amazing day. Doris H.