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Let $C_{1} : 9 x^{2} - 16 y^{2} - 18 x + 32 y - 23 = 0$ and $C_{2} : 25 x^{2} + 9 y^{2} - 50 x - 18 y + 33 = 0$ are two conics then which of the following options is/are correct?

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Area of the quadrilateral with vertices at the foci of the conics is $\frac{8}{9}$

Eccentricity of $C_{1} i s \frac{5}{4}$

Latus rectum of $C_{1}$ is greater than latus rectum of $C_{2}$

Eccentricity of $C_{2} i s \frac{5}{3}$

answer is A, C, D.

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Detailed Solution

$9 [x^{2} - 2 x] - 16 [y^{2} - 2 y] = 23$

$9 [{(x - 1)}^{2} - 1] - 16 [{(y - 1)}^{2} - 1] = 23$

$9 {(x - 1)}^{2} - 16 {(y - 1)}^{2} = 16$

$\frac{{(x - 1)}^{2}}{{(\frac{4}{3})}^{2}} - {(y - 1)}^{2} = 1$ which is a hyperbola

$C_{1} : \frac{{(x - 1)}^{2}}{{(\frac{4}{3})}^{2}} - {(y - 1)}^{2} = 1$

$C_{2} : \frac{{(x - 1)}^{2}}{{(\frac{1}{5})}^{2}} - \frac{{(y - 1)}^{2}}{{(\frac{1}{3})}^{2}} = 1$

$w i t h e_{E} = \frac{4}{5} [e_{E}^{2} = 1 - \frac{9}{25} = \frac{16}{25} \Rightarrow e_{E} = \frac{4}{5}]$

$a n d H : \frac{X^{2}}{(\frac{4}{3})} - Y^{2} = 1$

$f o c i o f e l l i p s e (0, \frac{4}{15}) a n d (0, \frac{- 4}{15})$

$f o c i o f h y p e r b o l a (\frac{5}{3}, 0) a n d (\frac{- 5}{3}, 0)$

$H e n c e a r e a = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} . \frac{8}{15} . \frac{10}{3} = \frac{8}{9}$

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