An equation of the tangent plane to the surface $16 z = 4 x ^ { 2 } + y ^ { 2 }$ at (2, 4, 2) is

A) $- 2 x + y - 2 z = 4$
B) $2 x - y + 2 z = 4$
C) $2 x + y + 2 z = 4$
D) $2 x + y - 2 z = 4$
E) $2 x - y - 2 z = 4$

Question

An equation of the tangent plane to the surface $16 z = 4 x ^ { 2 } + y ^ { 2 }$ at (2, 4, 2) is

A) $- 2 x + y - 2 z = 4$
B) $2 x - y + 2 z = 4$
C) $2 x + y + 2 z = 4$
D) $2 x + y - 2 z = 4$
E) $2 x - y - 2 z = 4$

Quizplus · Accepted Answer

To find the equation of the tangent plane to the given surface at a point, we first need to find the gradient of the surface at that point. The surface is given by $16z = 4x^2 + y^2$, which can be rewritten as $F(x, y, z) = 4x^2 + y^2 - 16z = 0$. The gradient of $F$, $\nabla F$, gives the normal vector to the surface at any point $(x, y, z)$, and is given by $\nabla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right)$.Calculating the partial derivatives, we get:- $\frac{\partial F}{\partial x} = 8x$- $\frac{\partial F}{\partial y} = 2y$- $\frac{\partial F}{\partial z} = -16$At the point (2, 4, 2), these derivatives are:- $\frac{\partial F}{\partial x} = 8(2) = 16$- $\frac{\partial F}{\partial y} = 2(4) = 8$- $\frac{\partial F}{\partial z} = -16$So, the normal vector to the surface at (2, 4, 2) is $N = (16, 8, -16)$. The equation of the tangent plane at a point $(x_0, y_0, z_0)$ with a normal vector $N = (a, b, c)$ is given by $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$. Substituting the values, we get $16(x - 2) + 8(y - 4) - 16(z - 2) = 0$, which simplifies to $16x + 8y - 16z = 32 + 32 - 32$, or $2x + y - 2z = 4$, which matches option D.

An Equation of the Tangent Plane to the Surface 16z=4x2+y216 z = 4 x ^ { 2 } + y ^ { 2 }16z=4x2+y2

An Equation of the Tangent Plane to the Surface $16 z = 4 x ^ { 2 } + y ^ { 2 }$