Question

Find the eqn of the hyperbola where foci are (0,+_12)and length of latus rectum is 36

Answer 1

there are two equations

hope this answer will help

Answer 2

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=x^{2}+3y^{2}=-108}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = (0,\pm12)} \\ \\ \tt{ : \implies Length\:of\:latus\:rectum = 36} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: -\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\\bold{As\:we\:know\:that}\\ \tt{:\implies foci=\sqrt{(12-(-12))^{2}+(0-0)^{2}}}\\\\ \tt{:\implies 2be=24}\\\\ \tt{: \implies be = 12}$

$\text{Both \: side \: squaring} \\ \tt{: \implies {b}^{2} {e}^{2} = 144} - - - - - (2) \\ \\ \bold{As\:we\:know\:that} \\ \tt{: \implies Length \: of \: latus \: rectum = 36 }\\ \\ \tt{: \implies \frac{2 {a}^{2} }{b} = 36} \\ \\ \tt{: \implies {a}^{2} = 18b - - - - - (3)} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies {a}^{2} = {b}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {a}^{2} = {b}^{2} {e}^{2} - {b}^{2} } \\ \\ \tt{: \implies {a}^{2} + {b}^{2} = {b}^{2} {e}^{2} } -----(4)\\ \\ \text{Putting \: values \: of \: (1) \: and \: (2) \: in \: (4)} \\ \tt{: \implies 18b+ {b}^{2} = 144} \\ \\ \tt{: \implies {b}^{2} + 18b - 144= 0} \\ \\ \tt{: \implies(b-6)(b + 24) = 0} \\ \\ \tt{: \implies b = 6 \:and \: - 24} \\ \\ \tt{: \implies b = 6} \: \: \: (Value \: of \: b \: not \: be \: negative) \\ \\ \bold{Putting \: value \: of \: b\: in \: (3)} \\ \tt{: \implies {a}^{2} = 18\times 6 } \\ \\ \green{\tt{ : \implies {a}^{2} = 108}}\\ \\ \text{Putting \: value \: of \: a \: and \: b \: in \: (1)} \\ \tt{:\implies \frac{x^{2}}{108}+\frac{y^{2}}{36}=-1 }\\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: {x}^{2} + 3{y}^{2} = -108}}$