Axes are co-ordinate axes, S and S¹ are foci, B and B¹ are the ends of minor axis $\overline{){\mathrm{SBS}}^{\mathrm{\prime }}}={\sin }^{-1}\left(\frac{4}{5}\right)$.If area of SB S¹B¹ is 20 sq. units then equation of the ellipse is

$$\begin{gathered} Let the ellipse be \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) \\ Given, area of SBS\text{'}B\text{'}=20 \\ \Rightarrow \frac{1}{2}\left(2ae\right)\times \left(2b\right)=20 \\ \Rightarrow aeb=10 \\ \Rightarrow a^2{e}^2=\frac{b^2}{100}\cdots \left(1\right) \\ And \sin \theta =\frac{4}{5} \\ \Rightarrow \tan \theta =\frac{4}{3} \\ \Rightarrow \frac{4}{3}=\frac{2\tan {\displaystyle \frac{\theta }{2}}}{1-{\tan }^2\frac{\theta }{2}} \\ \Rightarrow 2{\tan }^2\frac{\theta }{2}+3\tan \frac{\theta }{2}-2=0 \\ \Rightarrow (2\tan \frac{\theta }{2}-1)(\tan \frac{\theta }{2}+2)=0 \\ \Rightarrow \tan \frac{\theta }{2}=\frac{1}{2}(\because \tan \frac{\theta }{2}\ne -2,\theta \in (0,\pi \left)\right) \\ \Rightarrow \frac{ae}{b}=\frac{1}{2}\cdots \left(2\right) \\ From equations \left(1\right) and \left(2\right), we get b^2=20\Rightarrow a^2=25 \\ \therefore Equation of ellipse will be \frac{x^2}{25}+\frac{y^2}{20}=1 \end{gathered}$$