If the equation of directrix and it's corresponding focus of an ellipse are x+y=2 and (2,3). If the eccentricity be 1/2 then the length of the latus rectum is-​  

Dear Student,

Please find below the solution to the asked query :
Here is the hint for the same.
The focus of the ellipse is at F(2,3), the corresponding directrix is the line x +y-2 and e = 1/2.
Let P (x, y) be any point on the ellipse and | MP | be the perpendicular distance from P to the directrix, then by def. of ellipse
 |FP| = e |MP|
$$\begin{gathered} \sqrt{{\left(\mathrm{x}-2\right)}^2+{\left(\mathrm{y}-3\right)}^2}= \frac{1}{2}\times \frac{\left|\mathrm{x}+\mathrm{y}-2\right|}{\sqrt{1^2+1^2}}= \frac{1}{2\sqrt{2}}\left|\mathrm{x}+\mathrm{y}-2\right| \\ 8\left[{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+13\right]= {\mathrm{x}}^2+{\mathrm{y}}^2+4+2\mathrm{xy}-4\mathrm{y}-4\mathrm{x} \\ \mathrm{Now} \mathrm{represent} \mathrm{it} \mathrm{in} \mathrm{the} \mathrm{form} \mathrm{of} \\ \frac{{\left(\mathrm{X}-\mathrm{h}\right)}^2}{{\mathrm{a}}^2}+\frac{{\left(\mathrm{Y}-\mathrm{K}\right)}^2}{{\mathrm{b}}^2}= 1 \\ \mathrm{and} \mathrm{then} \mathrm{length} \mathrm{of} \mathrm{latus} \mathrm{rectum} = \frac{2{\mathrm{b}}^2}{\mathrm{a}} \end{gathered}$$
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