Question

Area bounded by the circle which is concentric with the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and which passes through
$(4,-\frac{9}{5})$, the vertical chord common to both circle and ellipse on the positive side of x-axis is

• A. $\frac{481}{25}ta{n}^{-1}\left(\frac{9}{20}\right)-\frac{36}{5}$
  
• B. $2ta{n}^{-1}\left(\frac{9}{20}\right)$
  
• C. $\frac{481}{25}ta{n}^{-1}\left(\frac{9}{20}\right)$
• D. None of these

Answer 1

The correct option is A $\frac{481}{25}ta{n}^{-1}\left(\frac{9}{20}\right)-\frac{36}{5}$




As eccentricity of ellipse = $\frac{4}{5}$
Co-ordinate of foci = (4, 0), (-4, 0)
$\Rightarrow (4-\frac{9}{5})$ is one of the end point of latus - rectum
$\Rightarrow$ required area is
$\frac{1}{2\pi }\times \pi \times (4^2+\frac{9^2}{5^2})\times 2ta{n}^{-1}\left(\frac{9}{20}\right)$ -area of $\Delta POQ$
$=\frac{481}{25}ta{n}^{-1}\left(\frac{9}{20}\right)-\frac{1}{2}\times 4\times \left(\frac{18}{5}\right)=\frac{481}{25}ta{n}^{-1}\left(\frac{9}{20}\right)-\frac{36}{5}$