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:
Show Hint
- \( \frac{7}{5} \)
- \( \frac{7\sqrt{2}}{5} \)
- \( \frac{14}{5} \)
- \( \frac{14\sqrt{2}}{5} \)
The Correct Option is B
Solution and Explanation
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: \[ \boxed{(B) \frac{7\sqrt{2}}{5}}. \]
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