Question

If β be the length of the latus rectum of the hyperbola 16x^2-9x^2+32x+36y-164=0 then3β is

Answer 1

CORRECT QUESTION :

If β be the length of the latus rectum of the hyperbola 16x^2-9y^2+32x+36y-164=0 then3β is.

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

$\green{\tt{\therefore{3\beta=39.21\:units}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: hyperbola= 16x^{2}-9y^{2}+36y+32x-164=0 } \\ \\ \tt{:\implies Latus\:rectum=\beta} \\\\ \red {\underline \bold{To \: Find: }}\\ \tt{:\implies Eccentricity(e)=?}\\\\ \tt{:\implies Foci=?}\\ \\ \tt {: \implies 3\beta=?}$

• According to given question :

$\tt {: \implies 16{x}^{2} - {9y}^{2} + 36y+32x -164= 0} \\ \\ \tt{: \implies 16{x}^{2} +32x + 16 - 16-9 {y}^{2} + 36y + 36 -36-164= 0} \\ \\ \tt{: \implies (4x + 4)^{2} -(3y + 6)^{2} = 216} \\ \\ \tt{ :\implies \frac{(x +1)^{2} }{\frac{216}{16}} - \frac{(y + 2) ^{2} }{ \frac{216}{9} } = 1 } \\ \\ {: \implies \frac{ {(x+ 1)}^{2} }{13.5} + \frac{ {(y + 2)}^{2} }{24} = 1}$

$\text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 13.5} \\ \\ \tt{\circ \: {b}^{2} = 24}$

$\bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 24 }{3.67} } \\ \\ \green{\tt{ : \implies Latus \: rectum = 13.07\:units}}\\\\ \tt{:\implies 3\beta=3\times 13.07}\\\\ \green{\tt{\therefore 3\beta=39.21\:units}}$