Question

A point moves so that the sum of squares of its distances from two intersecting lines is constant (given that the lines are neither perpendicular nor they make complimentary angle). Prove that its locus is an ellipse.

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{line}\mathrm{is}\mathrm{ax}+\mathrm{by}+\mathrm{c}=0\mathrm{and}\mathrm{mx}+\mathrm{ny}+\mathrm{p}=0\mathrm{and}\mathrm{the}\mathrm{fixed}\mathrm{point}\mathrm{is} \\ \mathrm{P}\left(\mathrm{h},\mathrm{k}\right) \\ \mathrm{Now}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}, \\ {\left(\frac{\mathrm{ah}+\mathrm{bk}+\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)}^2+{\left(\frac{\mathrm{mh}+\mathrm{nk}+\mathrm{p}}{\sqrt{{\mathrm{m}}^2+{\mathrm{n}}^2}}\right)}^2=\mathrm{k} \\ \mathrm{Now}\mathrm{simplify}\mathrm{it}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{answer}. \end{gathered}$$