Question

The encentricity of a hyperbola is 8 and Foci is (+-16,0). Then find the Eqn of hyperbola.​

Answer 1

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=63x^{2}-y^{2}=252}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (\pm 16,0)} \\ \\ \tt{ : \implies Eccentricity(e) = 8} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2ae } \\ \\ \tt{ : \implies Foci \ = \sqrt{(16-(-16))^{2}+(0-0)^{2}}=32} \\ \\ \tt{ : \implies 2ae = 32} \\ \\ \tt{: \implies a \times {8} = 16} \\ \\ \green{ \tt{: \implies a = 2 }}\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {b}^{2} = 2^{2}( 8^{2} - 1)} \\ \\ \green{ \tt{ : \implies {b}^{2} =4 \times 63} }\\ \\ \tt{: \implies {b}^{2} = 252} \\\\ \tt{:\implies Co-ordinate\:of\:centre=\frac{16+(-16)}{2},\frac{0+0}{2}}\\\\ \tt{:\implies Co-ordinate\:of\:centre=(0,0) }\\\\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies \frac{ {x}^{2} }{4} - \frac{ {y}^{2} }{252} = 1}} \\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: 63{x}^{2} - {y}^{2} = 252}}$