For the hyperbola frac{x^2}{9} - frac{y^2}{3} | vedclass

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(D) Given hyperbola: $\frac{x^2}{9} - \frac{y^2}{3} = 1$. Here $a^2 = 9$ and $b^2 = 3$.Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{

Vedclass · Accepted Answer

Both $(A)$ and $(B)$.

Vedclass · Answer

(D) Given hyperbola: $\frac{x^2}{9} - \frac{y^2}{3} = 1$. Here $a^2 = 9$ and $b^2 = 3$.Eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{3}{9}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$. Thus,statement $(B)$ is incorrect.Length of latus rectum $LLR = \frac{2b^2}{a} = \frac{2 	imes 3}{3} = 2$. Thus,statement $(C)$ is correct.The product of perpendicular distances from any point $(x_1, y_1)$ on the hyperbola to the asymptotes $bx \pm ay = 0$ is given by $p_1 p_2 = \frac{a^2 b^2}{a^2 + b^2} = \frac{9 	imes 3}{9 + 3} = \frac{27}{12} = \frac{9}{4} = 2.25$.Since $2.25 > 2$ $(LLR)$,the statement in $(A)$ is incorrect.Since both $(A)$ and $(B)$ are incorrect,the correct option is $(D)$.

For the hyperbola $\frac{x^2}{9} - \frac{y^2}{3} = 1$,the incorrect statement is:

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