Question

Find the equation of locus of a point such that, the difference of the squares of its distances from the point $(5,0)$ and $(2,3)$ is $10$.

Answer 1

$\begin{array}{c}Letpointbe\left(h,k\right)\\ dis\tan cefrompoint\left(5,0\right)and\left(2,3\right)is\\ Dis\tan ce=\sqrt{{\left(h-5\right)}^2+{\left(k-0\right)}^2}.......\left(1\right)\\ Dis\tan ce=\sqrt{{\left(h-2\right)}^2+{\left(k-3\right)}^2}.......\left(2\right)\\ squaringbothequationandsubstract\left(2\right)-\left(1\right)\\ {\left[\sqrt{{\left(h-2\right)}^2+{\left(k-3\right)}^2}\right]}^2-{\left[\sqrt{{\left(h-5\right)}^2+{\left(k-0\right)}^2}\right]}^2=10\\ {\left(h-2\right)}^2+{\left(k-3\right)}^2-\left[{\left(h-5\right)}^2+k^2\right]=10\\ h^2-2h+4+k^2-6k+9-\left[h^2-10h+25+k^2\right]=10\\ h^2-2h+4+k^2-6k+9-h^2+10h-25-k^2=10\\ 8h-6k-25=10\\ 8h-6k=35itisrequiredlocus\end{array}$