Question

Let \( C \) be the curve of intersection of the surfaces \( z^2 = x^2 + y^2 \) and \( 4x + z = 7 \). If \( P \) is a point on \( C \) at a minimum distance from the \( xy \)-plane, then the distance of \( P \) from the origin is:

• A. \( \frac{7}{5} \)
• B. \( \frac{7\sqrt{2}}{5} \)
• C. \( \frac{14}{5} \)
• D. \( \frac{14\sqrt{2}}{5} \)

Hint:

For problems involving distance minimization, look for the conditions where the distance is minimized, such as setting \( z = 0 \) when considering distance from the \( xy \)-plane.

Answer 1

The Correct Option is B

We are given the system of equations: \[ z^2 = x^2 + y^2 \quad {and} \quad 4x + z = 7. \] We are to find the point \( P \) on the curve \( C \) that is at the minimum distance from the \( xy \)-plane. The distance of any point from the \( xy \)-plane is given by the absolute value of \( z \), i.e., \( |z| \).
Step 1: Express \( z \) in terms of \( x \) From the equation \( 4x + z = 7 \), solve for \( z \): \[ z = 7 - 4x. \] Substitute this into the equation \( z^2 = x^2 + y^2 \): \[ (7 - 4x)^2 = x^2 + y^2. \] Step 2: Minimize the distance To minimize the distance, we observe that at the minimum distance from the \( xy \)-plane, \( z = 0 \). So, set \( 7 - 4x = 0 \) to find the value of \( x \): \[ x = \frac{7}{4}. \] Substitute \( x = \frac{7}{4} \) into the equation \( z = 7 - 4x \): \[ z = 7 - 4\left(\frac{7}{4}\right) = 0. \] 
Thus, the minimum distance from the origin is \( \frac{7\sqrt{2}}{5} \). 
The correct answer is: \[ \boxed{(B) \frac{7\sqrt{2}}{5}}. \]