I'm trying to solve this task:
Find an envelope to a family of planes $z-z_0=p(x-x_0)+q(y-y_0)$ where $p^2+q^2=1$
I tried finding partial derivatives with respect to $p$ and $q$ but all I'm getting is that $(x-x_0), (y-y_0), (z-z_0)$ should just all be zero.
Correct answer is:
$(z-z_0)^2=(x-x_0)^2+(y-y_0)^2$
but I'm not sure how to get it.
Could somebody please give a hint? Thanks in advance.
To simplify notation, let's rewrite the equation $z-z_0=p(x-x_0)+q(y-y_0)$ as $$ Z=pX+qY, \tag{1} $$ where $(X,Y,Z):=(x-x_0,y-y_0,z-z_0).$ Next, since $p^2+q^2=1$, let's write $(p,q)=(\cos\phi, \sin\phi)$, so $(1)$ becomes $$ Z=X\cos\phi + Y\sin\phi. \tag{2} $$ To find the envelope of the family $(2)$, we must eliminate $\phi$ between $(2)$ and $$ \frac{\partial}{\partial\phi}(Z-X\cos\phi-Y\sin\phi)=0 \implies X\sin\phi-Y\cos\phi=0. \tag{3} $$ Solving $(3)$ for $\phi$, we get $$ \tan\phi=\frac{Y}{X}\implies\begin{cases} \cos\phi=\frac{1}{\sec\phi}=\pm\frac{1}{\sqrt{1+\tan^2\phi}}=\pm\frac{X}{\sqrt{X^2+Y^2}}, \\ \sin\phi=\tan\phi\,\cos\phi=\pm\frac{Y}{\sqrt{X^2+Y^2}}. \end{cases} \tag{4} $$ Plugging $(4)$ into $(2)$, we finally obtain $$ Z=\pm\frac{X^2+Y^2}{\sqrt{X^2+Y^2}}=\pm\sqrt{X^2+Y^2} \implies Z^2=X^2+Y^2, \tag{5} $$ or $$ (z-z_0)^2=(x-x_0)^2+(y-y_0)^2. \tag{6} $$