Question

Obtain a partial Differential equation by eliminating arbitrary constants from z=(x-α) ^2 +(y-β) ^2?

A) 2z=p^2+q^2

B) 4z=p^2+q^2

C) 4z=p+q

D) z^2=p^2+q^2

Answer 1

Solution:

$z=(x-α) ^2 +(y-β) ^2\ ...(i) \\ \Rightarrow \dfrac{\partial z}{\partial x}=2(x-\alpha) \\ \Rightarrow p=2(x-\alpha) \\ \Rightarrow (x-\alpha)=\dfrac p2\ ...(ii)$

Next, $\dfrac{\partial z}{\partial y}=2(y-\beta)$

$\Rightarrow q=2(y-\beta) \\ \Rightarrow (y-\beta)=\dfrac q2 \ ...(iii)$

Put (ii) and (iii) in (i).

$z=(\dfrac p2) ^2 +(\dfrac q2) ^2 \\ \Rightarrow 4z=p^2+q^2$

Option B is correct.