Q) a right angled triangle ABC having a right angle at C , $CA=b$ and $CB=a$ move such that the angular points A and B slide along x-axis and y-axis axis respectively. Find locus of C ?
Let C=(h,k) and A lies on x-axis and B lies on y-axis , i assumed an angle $\theta$ between x-axis and side $CA$ then i got coords of A, B as
$$A=(h+b\cos \theta ,0) $$ $$B=(0,a\sin \theta -k)$$
Then i wrote equations of equations of CA and CB . Then i found out their slopes and i used $m_{CA}.m_{CB}=-1$. Then i got a huge second degree differential equations , and i stopped there .
Should i resume my method or how else should i begin a approach ?
Let $C=(h,k)$, $A=(p,0)$, $B=(0,q)$.
Slope of AC=$\frac{k}{h-p}$
Slope of BC=$\frac{k-q}{h}$
Using distance formula, $$a=\sqrt{(h-p)^2+k^2}$$ $$h-p=\sqrt{a^2-k^2}$$ Similarly, $$k-q=\sqrt{b^2-h^2}$$
Substituting the above values on the slope equation,
Slope of AC=$\frac{k}{\sqrt{a^2-k^2}}$
Slope of BC=$\frac{\sqrt{b^2-h^2}}{h}$
Since AB is perpendicular to BC,$$\frac{k}{\sqrt{a^2-k^2}}\cdot \frac{\sqrt{b^2-h^2}}{h}=-1$$
Squaring both sides you get $$\frac{k^2}{a^2-k^2}\cdot \frac{b^2-h^2}{h^2}=1$$
$$(b^2-h^2)k^2=(a^2-k^2)h^2$$ $$bk \pm ah=0$$
Hence locus of Point C is $$by \pm ax=0$$