The locus of mid points of the chords of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ which pass through  the foot of a directrix 

Given ellipse$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Let mid point of chord be  $({\mathrm{x}}_1,{\mathrm{y}}_1)$ 

Now equation  of chord is  ${\mathrm{S}}_1={\mathrm{S}}_{11}$

$$\begin{gathered} \frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}+\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=\frac{{{\mathrm{x}}_1}^2}{{\mathrm{a}}^2}+\frac{{{\mathrm{y}}_1}^2}{{\mathrm{b}}^2} \\ \sin ce it passes through the point (\frac{a}{e},0) then \\ \frac{\left({\displaystyle \frac{a}{e}}\right)x_1}{a^2}+0=\frac{{{\mathrm{x}}_1}^2}{{\mathrm{a}}^2}+\frac{{{\mathrm{y}}_1}^2}{{\mathrm{b}}^2} \\ \therefore \mathrm{Locus} \mathrm{of} ({\mathrm{x}}_1 {\mathrm{y}}_1) \mathrm{is} \\ \frac{\mathrm{x}}{\mathrm{ae}}=\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} \end{gathered}$$