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The equation of the hyperbola whose foci are (6, 4) and (- 4, 4) and eccentricity 2 is given by

Option: 1

\mathrm{12 x^2-4 y^2-24 x+32 y-127=0}


Option: 2

\mathrm{12 x^2+4 y^2+24 x-32 y-127=0}


Option: 3

\mathrm{12 x^2-4 y^2-24 x-32 y+127=0}


Option: 4

\mathrm{12 x^2-4 y^2+24 x+32 y+127=0}


Answers (1)

best_answer

Foci are (6,4) and (-4,4) and \mathrm{e=2.}
\mathrm{\therefore \text { Centre is }\left(\frac{6-4}{2}, \frac{4+4}{2}\right)=(1,4)}
\mathrm{\text { So, } a e+1=6 \Rightarrow a e=5 \Rightarrow a=\frac{5}{2} \text { and } b=\frac{5}{2} \sqrt{3}}
Hence, the required equation is \mathrm{\frac{(x-1)^2}{25 / 4}-\frac{(y-4)^2}{(75 / 4)}=1 \text { or } 12 x^2-4 y^2-24 x+32 y-127=0}

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manish painkra

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