The symmetric equations of the tangent line to the curve of intersection of the surface $z = x ^ { 2 } + y ^ { 2 }$ and the plane y = 2 at the point (1, 2, 5) are

A) $z - 5 = 2 ( x - 1 ) , y = 2$
B) $z - 5 = \frac { x - 1 } { - 2 } , y = 2$
C) $2 z - 5 = \frac { x - 1 } { 2 } , y = 2$
D) $2 z - 5 = \frac { x - 1 } { - 2 } , y = 2$
E) $4 z - 5 = \frac { x - 1 } { 2 } , y = 2$

Question

The symmetric equations of the tangent line to the curve of intersection of the surface $z = x ^ { 2 } + y ^ { 2 }$ and the plane y = 2 at the point (1, 2, 5) are

A) $z - 5 = 2 ( x - 1 ) , y = 2$
B) $z - 5 = \frac { x - 1 } { - 2 } , y = 2$
C) $2 z - 5 = \frac { x - 1 } { 2 } , y = 2$
D) $2 z - 5 = \frac { x - 1 } { - 2 } , y = 2$
E) $4 z - 5 = \frac { x - 1 } { 2 } , y = 2$

Quizplus · Accepted Answer

To find the tangent line at the point (1, 2, 5) on the curve of intersection, we first need the gradient of the surface $z = x^2 + y^2$. The gradient vector is $\nabla z = (2x, 2y, -1)$ at any point (x, y, z). At the point (1, 2, 5), this becomes (2, 4, -1). Since the plane is y = 2, the direction of the tangent line is parallel to the xz-plane, and its direction vector can be taken from the x and z components of the gradient, which are 2 and -1, respectively. Thus, the symmetric equations of the tangent line are $z - 5 = 2(x - 1)$ and $y = 2$, matching option A.

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

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