Question

Find the length of latus rectum of hyperbola x^2/36 - y^2/16 =1​

Answer 1

$\large\underline{\sf{Given- }}$

$\sf \: The \: equation \: of \: hyperbola : \: \dfrac{ {x}^{2} }{36} - \dfrac{ {y}^{2} }{16} = 1$

$\large\underline{\sf{To\:Find - }}$

$\: \: \: \: \: \: \: \: \: \: \bull \: \: \: \: \sf \: Length \: of \: Latus \: Rectum$

$\large\underline{\sf{Solution-}}$

The given equation of Hyperbola is

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{36} - \dfrac{ {y}^{2} }{16} = 1$

On comparing with

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

We get

$\rm :\longmapsto\: {a}^{2} = 36\rm :\implies\:a = 6$

$\rm :\longmapsto\: {b}^{2} = 16\rm :\implies\:b = 4$

We know,

$\rm :\longmapsto\:Length \: of \: Latus \: Rectum \: = \: \dfrac{ {2b}^{2} }{a}$

$\rm :\longmapsto\:Length \: of \: Latus \: Rectum = \dfrac{2 \times 16}{6}$

$\rm :\longmapsto\:Length \: of \: Latus \: Rectum = \dfrac{16}{3}$

Additional Information :-

$\rm :\longmapsto\:The \: {eq}^{n} \: of \: hyperbola \: : \: \dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

then

$\boxed{ \bf{ \: Vertex ( \pm \: a, \: 0)}}$

$\boxed{ \bf{ \: eccentricity \: (e) = \sqrt{1 + \dfrac{ {b}^{2} }{ {a}^{2} } } }}$

$\boxed{ \bf{ \: Focus \: ( \pm \: ae,0)}}$

$\boxed{ \bf{ \: Length \: of \: transverse \: axis = 2a}}$

$\boxed{ \bf{ \: Length \: of \: conjugate \: axis = 2b}}$

Answer 2

Answer:

$The \: equation \: of \: hyperbola \: :36x2−16y2=1 \\ </p><p></p><p>\large\underline{\sf{To\:Find - }}ToFind− \\ </p><p></p><p>\: \: \: \: \: \: \: \: \: \: \bull \: \: \: \: \sf \: Length \: of \: Latus \: Rectum∙LengthofLatusRectum \\ </p><p></p><p>\large\underline{\sf{Solution-}}Solution− \\ </p><p></p><p>The \: given \: equation \: of \: Hyperbola \: is \\ </p><p></p><p>\rm :\longmapsto\:\dfrac{ {x}^{2} }{36} - \dfrac{ {y}^{2} }{16} = 1:⟼36x2−16y2=1 \\ </p><p></p><p>On \: comparing \: with \\ </p><p></p><p>\rm :\longmapsto\:\dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1:⟼a2x2−b2y2=1 \\ </p><p></p><p>We \: get \\ </p><p></p><p>\rm :\longmapsto\: {a}^{2} = 36\rm :\implies\:a = 6:⟼a2=36:⟹a=6 \\ </p><p></p><p>\rm :\longmapsto\: {b}^{2} = 16\rm :\implies\:b = 4:⟼b2=16:⟹b=4 \\ </p><p></p><p>We know, \\ </p><p></p><p>\rm :\longmapsto\:Length \: of \: Latus \: Rectum \: = \: \dfrac{ {2b}^{2} }{a}:⟼LengthofLatusRectum=a2b2 \\ </p><p></p><p>\rm :\longmapsto\:Length \: of \: Latus \: Rectum = \dfrac{2 \times 16}{6}:⟼LengthofLatusRectum=62×16 \\ </p><p></p><p>\rm :\longmapsto\:Length \: of \: Latus \: Rectum = \dfrac{16}{3}:⟼LengthofLatusRectum=316 \\ </p><p></p><p>Additional \: Information :- \\ </p><p></p><p>\rm :\longmapsto\:The \: {eq}^{n} \: of \: hyperbola \: : \: \dfrac{ {x}^{2} }{ {a}^{2} } - \dfrac{ {y}^{2} }{ {b}^{2} } = 1:⟼Theeqnofhyperbola:a2x2−b2y2=1 \\ </p><p></p><p>then \\ </p><p></p><p>\boxed{ \bf{ \: Vertex ( \pm \: a, \: 0)}}Vertex(±a,0) \\ </p><p></p><p>\boxed{ \bf{ \: eccentricity \: (e) = \sqrt{1 + \dfrac{ {b}^{2} }{ {a}^{2} } } }}eccentricity(e)=1+a2b2 \\ </p><p></p><p>\boxed{ \bf{ \: Focus \: ( \pm \: ae,0)}}Focus(±ae,0) \\ </p><p></p><p>\boxed{ \bf{ \: Length \: of \: transverse \: axis = 2a}}Lengthoftransverseaxis=2a \\ </p><p></p><p>\boxed{ \bf{ \: Length \: of \: conjugate \: axis = 2b}}Lengthofconjugateaxis=2b \\ \\ </p><p></p><p>$