for a variable line $\frac xa +\frac yb =1$, find the locus of foot of perpendicular drawn from origin to the line under the condition that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
The line perpendicular to this line and passing through origin is $y=\frac ab x$
The point of interaction of this line is $$x=\frac{ab^2}{a^2+b^2}$$ and $$y=\frac{a^2b}{a^2+b^2}$$
While $c^2 =\frac{a^2b^2}{a^2+b^2}$
So $ax=c^2$ and $by=c^2$
How should I proceed from here?
Notice that for $x = \dfrac {ab^2}{a^2+b^2}$, $y = \dfrac {a^2b}{a^2+b^2}$, we have:
$$x^2 + y^2 = \frac {(ab^2)^2 + (a^2b)^2}{(a^2+b^2)^2} = \frac {a^2b^2(a^2+b^2)}{(a^2+b^2)^2} = \frac {a^2b^2}{a^2+b^2} = \left(\frac1{a^2} + \frac1{b^2}\right)^{-1} = c^2$$
What do you think this shape is?
Alternatively, the line $\dfrac xa + \dfrac yb = 1$ passes through the points $(a,0)$ and $(0,b)$. The length of the perpendicular from the origin to this line can be calculated by considering the area of the triangle with $(0,0), (a,0), (0,b)$ as vertices. We would therefore have:
$$\frac{ab}2 = \frac {h\sqrt{a^2+b^2}}2$$
giving $h = \dfrac{ab}{\sqrt{a^2+b^2}}$, or $h^2 = \dfrac {a^2b^2}{a^2+b^2} = c^2 \leadsto h = |c|$.
Hint
WLOG $a=c\cos t,b=c\sin t$
$$x/c\cos t+y/c\sin t=1$$
so, the equation of the perpendicular will be $$x/\sin t-y/\cos t=0$$
Solve for $x,y$
and eliminate$t$
