Question

The partial differential equation obtained by eliminating the constants \(a\) and \(b\) from \[ z = (x - a)^2 + (y - b)^2 \] is:

• A. \(4z = \left( \dfrac{\partial z}{\partial x} \right)^2 + \left( \dfrac{\partial z}{\partial y} \right)^2\)
• B. \(4z = \dfrac{\partial z}{\partial x} + \dfrac{\partial z}{\partial y}\)
• C. \(4z = \left( \dfrac{\partial z}{\partial x} \right)^2 + \left( \dfrac{\partial z}{\partial y} \right)^2 + 1\)
• D. \(4z = \left( \dfrac{\partial z}{\partial x} \right)^2 - \left( \dfrac{\partial z}{\partial y} \right)^2\)

Hint:

To eliminate constants from a PDE, use partial derivatives and substitute back into the original equation.

Answer 1

The Correct Option is A

The equation given is:
\[z = (x - a)^2 + (y - b)^2\]

To eliminate the constants \(a\) and \(b\), we differentiate with respect to \(x\) and \(y\).

Step 1: Differentiate with respect to \(x\):
\[ \frac{\partial z}{\partial x} = 2(x - a) \]

Step 2: Differentiate with respect to \(y\):
\[ \frac{\partial z}{\partial y} = 2(y - b) \]

Square both partial derivatives and add them:
\[ \left( \frac{\partial z}{\partial x} \right)^2 = 4(x - a)^2,\quad \left( \frac{\partial z}{\partial y} \right)^2 = 4(y - b)^2 \]

Adding these results in:
\[ \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 = 4(x - a)^2 + 4(y - b)^2 \]

Recognize the right-hand side as \(4z\):
\[ 4z = (x - a)^2 + (y - b)^2 = \frac{1}{4}\left(\left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2\right) \]

Thus, the partial differential equation becomes:
\[ 4z = \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 \]

This matches the first option, confirming it as the correct choice.