Question

The differential equation of (x-a)^{2}+(y-b)^{2}+z^{2}=2019 where a, b are arbitrary constants is.

A) z^2(p^2+q^2)=1. B)z^2(p^2+q^2)=2019.

C) z^2(p^2+q^2+1)=1.

D) z^2(p^2+q^2+1)=2019

Answer 1

$$z=(x-a)^2+(y-b)^2$$

Differentiating partially with respect to $x$ and $y,$ we get

$$\dfrac{\partial z}{\partial x}=2(x-a)$$ $$\dfrac{\partial z}{\partial y}=2(y-b)$$

Squaring and adding these equations, we have

$$(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=(2(x-a))^2+(2(y-b))^2$$ $$(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=4((x-a)^2+(y-b)^2)$$

Since $(x-a)^2+(y-b)^2=z,$ we get

$$(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=4z$$