Question

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)

Answer 1

To find the parametric equations for the tangent line at a given point to an ellipse that lies on a plane, we use the gradients at that point to form the direction vector of the tangent line. For the ellipse formed by the intersection of ellipsoid 4x²+2y²+z²=16 and the plane y=2, the tangent line equations at point (1,2,2) are x=1+8t, y=2, z=2+4t.

Step-by-step explanation:

To find the parametric equations for the tangent line to the ellipse created by the intersection of the ellipsoid 4x²+2y²+z²=16 and the plane y=2 at the point (1,2,2), we first substitute y=2 into the ellipsoid equation to get the equation of the ellipse on that plane, which simplifies to 4x²+z²=12. To find the tangent line at the specific point, we need the gradient of the ellipse at that point. This can be found by taking the partial derivatives of the ellipse equation with respect to x and z.

The gradients are:

• ∂/∂x (4x²+z²) = 8x
• ∂/∂z (4x²+z²) = 2z

At the point (1,2,2), these gradients are (8,4). Now to find the tangent line, we use the point-gradient formula where the gradients form a direction vector. Hence, the parametric equations of the tangent line are:

• x = 1 + 8t
• y = 2
• z = 2 + 4t

The parameter

t

can be any real number.