The ellipsoid $8 x ^ { 2 } + 3 y ^ { 2 } + z ^ { 2 } = 14$ intersects the plane $y = 2$ in an ellipse. Find parametric equations for the tangent line to this ellipse at the point $( 1,2,2 )$.
A) $x = 1 + t , y = 2 , z = - 4 t + 2$
B) $x = 1 + t , y = 2 , z = - 4 t - 2$
C) $x = 1 + t , y = 1 , z = 4 t + 2$
D) $x = t , y = 2 , z = - 4 t + 2$
E) $x = 1 + t , y = 2 , z = 4 t - 2$

Question

Quizplus · Accepted Answer

To find the tangent line at the point (1,2,2) on the ellipse formed by the intersection of the ellipsoid $8x^2 + 3y^2 + z^2 = 14$ and the plane $y=2$, we substitute $y=2$ into the ellipsoid equation to get the equation of the ellipse in the $xz$-plane. This gives $8x^2 + 3(2)^2 + z^2 = 14$ or $8x^2 + z^2 = 2$. To find the tangent line, we need the gradient of the ellipsoid at the point (1,2,2), which can be found by taking the partial derivatives of the ellipsoid equation: $\nabla f = (16x, 6y, 2z)$. At the point (1,2,2), this gradient is $(16, 12, 4)$. However, since we're interested in the direction of the tangent in the $xz$-plane (because $y$ is constant), we focus on the $x$ and $z$ components, leading to a direction vector for the tangent line of $(1, -4)$ in the $xz$-plane, hence the parametric equations $x = 1 + t$, $y = 2$, $z = -4t + 2$.

The Ellipsoid 8x2+3y2+z2=148 x ^ { 2 } + 3 y ^ { 2 } + z ^ { 2 } = 148x2+3y2+z2=14

The Ellipsoid $8 x ^ { 2 } + 3 y ^ { 2 } + z ^ { 2 } = 14$