Question

If the equation of the tangent plane to the surface \( z = 16 - x^2 - y^2 \) at the point \( P(1, 3, 6) \) is \( ax + by + cz + d = 0 \), then the value of \( |d| \) is

• A. 16
• B. 26
• C. 36
• D. 46

Hint:

For a surface \( z = f(x, y) \), the tangent plane at \( (x_0, y_0, z_0) \) is \( z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \).

Answer 1

The Correct Option is B

Step 1: Find the partial derivatives.
Given \( z = 16 - x^2 - y^2 \), we have
\( \dfrac{\partial z}{\partial x} = -2x, \dfrac{\partial z}{\partial y} = -2y. \)

Step 2: Equation of tangent plane.
At any point \( (x_0, y_0, z_0) \), the tangent plane to \( z = f(x, y) \) is given by
\( z - z_0 = f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0). \)

Step 3: Substitute given point \( P(1, 3, 6) \).
\( f_x(1, 3) = -2(1) = -2, f_y(1, 3) = -2(3) = -6. \)
Thus, the tangent plane is
\( z - 6 = -2(x - 1) - 6(y - 3). \)

Step 4: Simplify.
\( z - 6 = -2x + 2 - 6y + 18 $\Rightarrow$ 2x + 6y + z - 26 = 0. \)

Step 5: Identify coefficients.
Here, \( a = 2, \, b = 6, \, c = 1, \, d = -26. \) Therefore, \( |d| = 26. \)

Final Answer: \( |d| = 26. \)