14th one

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ 9{\mathrm{x}}^2+25{\mathrm{y}}^2=225 \\ \Rightarrow \frac{9{\mathrm{x}}^2}{225}+\frac{25{\mathrm{y}}^2}{225}=1 \\ \Rightarrow \frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{9}=1...\left(\mathrm{i}\right) \\ \mathrm{Standard} \mathrm{equation} \mathrm{of} \mathrm{ellipse} \mathrm{is} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ {\mathrm{a}}^2=25\Rightarrow \mathrm{a}=5 \\ {\mathrm{b}}^2=9\Rightarrow \mathrm{b}=3 \\ \mathrm{Ends} \mathrm{of} \mathrm{latus} \mathrm{ractum} \mathrm{are} \left(\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right),\left(\mathrm{ae},-\frac{{\mathrm{b}}^2}{\mathrm{a}}\right),\left(-\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right),\left(-\mathrm{ae},-\frac{{\mathrm{b}}^2}{\mathrm{a}}\right) \\ \mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{25-9}{25}}=\frac{4}{5} \\ \left(\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)=\left(5\times \frac{4}{5},\frac{9}{5}\right)=\left(4,\frac{9}{5}\right) \\ \left(\mathrm{ae},-\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)=\left(5\times \frac{4}{5},-\frac{9}{5}\right)=\left(4,-\frac{9}{5}\right) \\ \left(-\mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)=\left(-5\times \frac{4}{5},\frac{9}{5}\right)=\left(-4,\frac{9}{5}\right) \\ \left(-\mathrm{ae},-\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)=\left(-5\times \frac{4}{5},-\frac{9}{5}\right)=\left(-4,-\frac{9}{5}\right) \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{to} \left(\mathrm{i}\right) \mathrm{at} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{is} \\ \frac{{\mathrm{xx}}_1}{25}+\frac{{\mathrm{yy}}_1}{16}=1 \\ \mathrm{Now} \mathrm{just} \mathrm{put} \mathrm{above} \mathrm{four} \mathrm{points} \mathrm{in} \mathrm{place} \mathrm{of} \left({\mathrm{x}}_1,{\mathrm{y}}_1\right) \mathrm{to} \mathrm{get} \\ \mathrm{four} \mathrm{equations} \mathrm{of} \mathrm{tangents}. \end{gathered}$$

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