Question

Find the eqn of hyperbola whose foci are (6,4) anf (-4,4) and eccentricity is 2.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Eqn\:of\:hyperbola=12(x-1)^{2}-4(y-4)^{2}=75}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (6,4)\:and\:(-4,4)} \\ \\ \tt{ : \implies Eccentricity(e) = 2} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2ae } \\ \\ \tt{ : \implies Foci \ = \sqrt{(6-(-4))^{2}+(4-4)^{2}}=10} \\ \\ \tt{ : \implies 2ae = 10} \\ \\ \tt{: \implies a \times {2} = 5} \\ \\ \green{ \tt{: \implies a = \frac{5}{2} }}\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {b}^{2} = \frac{5}{2}^{2}( 2^{2} - 1)} \\ \\ \green{ \tt{ : \implies {b}^{2} =\frac{25}{4}\times 3} }\\ \\ \tt{: \implies {b}^{2} =\frac{75}{4}} \\\\ \tt{:\implies Co-ordinate\:of\:centre=\frac{6+(-4)}{2},\frac{4+4}{2}}\\\\ \tt{:\implies Co-ordinate\:of\:centre=(1,4) }\\\\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies \frac{ {(x-1)}^{2} }{\frac{25}{4}} - \frac{ {(y-4)}^{2} }{\frac{75}{4}} = 1}} \\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: 12{(x-1)}^{2} - 4{(y-4)}^{2} = 75}}$

Answer 2

Find the eqn of hyperbola whose foci are (6,4) anf (-4,4) and eccentricity is 2.