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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas : $ 16x^2-9y^2=576$

$\begin {array} {1 1} (A)\;Foci ( \pm 10, 0), Vertices (\pm 6, 0), e = 5/3, Length \: of \: the \: latus \: rectum = 64/3 \\ (B)\;Foci ( 0, \pm 10), Vertices (0, \pm 6), e = 5/3, Length \: of \: the \: latus \: rectum = 64/3 \\ (C)\;Foci ( \pm 5, 0), Vertices (\pm 6, 0), e = 5/3, Length \: of \: the \: latus \: rectum = 64/3 \\ (D)\;Foci (0, \pm 5), Vertices (0, \pm 6), e = 5/3, Length \: of \: the \: latus \: rectum = 64/3 \end {array}$

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  • Standard equation of a hyperbola where transverse axis is along x - axis is $ \large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2} $$=1$
  • Standard equation of a hyperbola whose transverse axis is along y - axis is $ \large\frac{y^2}{a^2} $$ - \large\frac{x^2}{b^2}$$=1$
  • $c^2=a^2+b^2$, where $c$ is the foci
  • Eccentricity $e = \large\frac{c}{a}$
  • Length of the latus rectum $ = \large\frac{2b^2}{a}$
Step 1 :
The given equation is $ 16x^2-9y^2=576$
Dividing throughout by 576 we get,
$ \large\frac{x^2}{36}$$-\large\frac{y^2}{64}$$=1$
Comparing the equation with the standard equation of parabola $ \large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$ we get,
$a^2=36 $ and $ b^2=64$
$ \therefore c^2=a^2+b^2$
(i.e) $c^2=36+64=100$
$ c = \sqrt{100}=10$
Step 2 :
Hence the coordinates of foci are : $( \pm 10, 0)$
The coordinates of vertices are $( \pm 6, 0)$
Eccentricity $ e = \large\frac{10}{6}$$ = \large\frac{5}{3}$
Length of the latus rectum is $ \large\frac{2b^2}{a}$$ = \large\frac{2 \times 64}{6}$
$ = \large\frac{64}{3}$
answered Jul 19, 2014 by thanvigandhi_1
 

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