Let 16 x^{2}-3 y^{2}-32 x-12 y=44 represent a | vedclass

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(A, B, C) The given equation of the hyperbola is $16 x^{2}-3 y^{2}-32 x-12 y=44$.Rearranging the terms,we get $16(x^{2}-2x)-3(y^{2}+4y)=44$.Completing

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eccentricity is $\sqrt{19 / 3}$

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(A, B, C) The given equation of the hyperbola is $16 x^{2}-3 y^{2}-32 x-12 y=44$.Rearranging the terms,we get $16(x^{2}-2x)-3(y^{2}+4y)=44$.Completing the square,$16(x-1)^{2}-16-3(y+2)^{2}+12=44$.$16(x-1)^{2}-3(y+2)^{2}=48$.Dividing by $48$,we get $\frac{(x-1)^{2}}{3}-\frac{(y+2)^{2}}{16}=1$.Comparing with the standard form $\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1$,we have $a^{2}=3 \Rightarrow a=\sqrt{3}$ and $b^{2}=16 \Rightarrow b=4$.Length of the transverse axis $= 2a = 2\sqrt{3}$. (Option $A$ is correct)Length of latus rectum $= \frac{2b^{2}}{a} = \frac{2 	imes 16}{\sqrt{3}} = \frac{32}{\sqrt{3}}$. (Option $B$ is correct)Eccentricity $e = \sqrt{1+\frac{b^{2}}{a^{2}}} = \sqrt{1+\frac{16}{3}} = \sqrt{\frac{19}{3}}$. (Option $C$ is correct)Equation of directrices: $x-h = \pm \frac{a}{e}$ $\Rightarrow x-1 = \pm \frac{\sqrt{3}}{\sqrt{19/3}} = \pm \frac{3}{\sqrt{19}}$ $\Rightarrow x = 1 \pm \frac{3}{\sqrt{19}}$. (Option $D$ is incorrect).

Let $16 x^{2}-3 y^{2}-32 x-12 y=44$ represent a hyperbola. Then,

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