The symmetric equations of the tangent line to the curve of intersection of the surface $z = 16 - x ^ { 2 } - y ^ { 2 }$ and the plane y = 2 at the point (1, 2, 11) are

A) $z - 11 = - 2 ( x - 1 ) , y = 2$
B) $z - 11 = \frac { x - 1 } { \frac { 1 } { 2 } } , y = 2$
C) $2 z - 11 = \frac { x - 1 } { - \frac { 1 } { 2 } } , y = 2$
D) $2 z - 11 = \frac { x - 1 } { \frac { 1 } { 2 } } , y = 2$
E) $4 z - 11 = \frac { x - 1 } { - \frac { 1 } { 2 } } , y = 2$

Question

The symmetric equations of the tangent line to the curve of intersection of the surface $z = 16 - x ^ { 2 } - y ^ { 2 }$ and the plane y = 2 at the point (1, 2, 11) are

A) $z - 11 = - 2 ( x - 1 ) , y = 2$
B) $z - 11 = \frac { x - 1 } { \frac { 1 } { 2 } } , y = 2$
C) $2 z - 11 = \frac { x - 1 } { - \frac { 1 } { 2 } } , y = 2$
D) $2 z - 11 = \frac { x - 1 } { \frac { 1 } { 2 } } , y = 2$
E) $4 z - 11 = \frac { x - 1 } { - \frac { 1 } { 2 } } , y = 2$

Quizplus · Accepted Answer

To find the tangent line, we first need the gradient of the surface $z = 16 - x^2 - y^2$ at the point (1, 2, 11). The gradient is $\nabla z = (-2x, -2y, 1)$ at any point on the surface. At (1, 2, 11), this becomes $(-2, -4, 1)$. Since y is constant (y=2), the tangent line in the xz-plane has a direction vector proportional to (-2, 1), leading to the symmetric equation $z - 11 = -2(x - 1)$, with $y = 2$ being the plane of intersection.

The Symmetric Equations of the Tangent Line to the Curve z=16−x2−y2z = 16 - x ^ { 2 } - y ^ { 2 }z=16−x2−y2

The Symmetric Equations of the Tangent Line to the Curve $z = 16 - x ^ { 2 } - y ^ { 2 }$