Mathematics - Ellipse Question with Solution | TestHub

Let point $\mathrm{A}(a,0)$ is on $x$-axis and $\mathrm{B}$ $(0,b)$ is on $y$-axis.
Let $\mathrm{P}(h,k)$ divides $\mathrm{AB}$ in the ratio $1:2$. So, by section formula $$\begin{array}{rl}& h=\frac{2(0)+1(a)}{1+2}=\frac{a}{3}\\ & k=\frac{2(b)+1(0)}{3}=\frac{2b}{3}\\ & \Rightarrow a=3h\text{ and }b=\frac{3k}{2}\end{array}$$ Now, $a^2+b^2=l^2$ $$\begin{array}{rl}& \Rightarrow 9h^2+\frac{9k^2}{4}=l^2\\ & \Rightarrow \frac{h^2}{{\left(\frac{l}{3}\right)}^2}+\frac{k^2}{{\left(\frac{2l}{3}\right)}^2}=1\end{array}$$ Now $\mathrm{e}=\sqrt{1-\left(\frac{l^2}{9}\times \frac{9}{4l^2}\right)}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$ Thus, required locus of $\mathrm{P}$ is an ellipse with eccentricity $\frac{\sqrt{3}}{2}$.