A point moves such that the sum of the square of its distances from two fixed straight lines intersecting at angle 2$\mathrm{\alpha }$ is a constant. The locus of point is an ellipse of eccentricity

Let us choose the point of intersection of the given lines as the origin and their angular bisector as the x-axis. Then equation of the two lines will be y = mx and y = -mx, where m = tan$\mathrm{\alpha }$.

Let P(h, k)be the point whose locus is to be found. Then according to the given condition,
PA² + PB² = Constant
$$\begin{gathered} \Rightarrow \frac{(\mathrm{k}-\mathrm{mh}{)}^2}{1+{\mathrm{m}}^2}+\frac{(\mathrm{k}+\mathrm{mh}{)}^2}{1+{\mathrm{m}}^2}=\mathrm{c} (\mathrm{c} \mathrm{is} \mathrm{a} \mathrm{constant}) \\ \Rightarrow 2\left({\mathrm{k}}^2+{\mathrm{m}}^2{\mathrm{h}}^2\right)=\mathrm{c}\left(1+{\mathrm{m}}^2\right) \end{gathered}$$
Therefore, the locus of point P is
$$\begin{gathered} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1, \\ \text{ where }{\mathrm{a}}^2=\frac{\mathrm{c}\left(1+{\mathrm{m}}^2\right)}{2{\mathrm{m}}^2}\text{ and }{\mathrm{b}}^2=\frac{\mathrm{c}\left(1+{\mathrm{m}}^2\right)}{2}. \\ \text{ If }\mathrm{\alpha }<\mathrm{\pi }/4\text{, then }\mathrm{m}<1\text{, and }{\mathrm{a}}^2>{\mathrm{b}}^2. \\ \therefore \text{Eccentricity }=\sqrt{1-\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{1-{\mathrm{m}}^2}=\frac{\sqrt{\cos 2\mathrm{\alpha }}}{\cos \mathrm{\alpha }} \\ \text{ If }\mathrm{\alpha }>\mathrm{\pi }/4\text{, then }\mathrm{m}>1\text{, and }{\mathrm{a}}^2<{\mathrm{b}}^2. \\ \therefore \text{Eccentricity }=\sqrt{1-\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}=\sqrt{1-\frac{1}{{\mathrm{m}}^2}}=\frac{\sqrt{-\cos 2\mathrm{\alpha }}}{\sin \mathrm{\alpha }} \end{gathered}$$