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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 : $ \large\frac{x^2}{16}-\large\frac{y^2}{9}=1$

$\begin {array} {1 1} (A)\;Foci ( \pm 5, 0), Vertices ( \pm 4, 0), eccentricity (e) : 5/4, Length\: of \: the \: latus \: rectum : 9/2\\ (B)\;Foci ( 0,\pm 5), Vertices ( \pm 4, 0), eccentricity (e) : 5/4, Length\: of \: the \: latus \: rectum : 9/2 \\ (C)\;Foci ( \pm 5, 0), Vertices (0, \pm 4), eccentricity (e) : 5/4, Length\: of \: the \: latus \: rectum : 9/2 \\ (D)\;Foci ( \pm 5, 0), Vertices ( \pm 4, 0), eccentricity (e) : 5/4, Length\: of \: the \: latus \: rectum : 2 \end {array}$

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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 of the hyperbola is $ \large\frac{x^2}{16}$$-\large\frac{y^2}{9}$$=1$
Comparing this with the standard equation $ \large\frac{x^2}{a^2} $$ - \large\frac{y^2}{b^2}$$=1$ we get,
$a^2=16 \Rightarrow a = 4$
$b^2=9 \Rightarrow b = 3$
$ \therefore c^2=a^2+b^2$
(i.e) $c^2 = 16+9$
$c = \sqrt{25}=5$
Step 2 :
The coordinates of foci are $( \pm 5, 0)$
The coordinates of vertices are $( \pm 4, 0)$
Eccentricity $ e = \large\frac{c}{a}$$= \large\frac{5}{4}$
Length of the latus rectum $ = \large\frac{2b^2}{a}$$ = \large\frac{2 \times 9}{4}$$ = \large\frac{9}{2}$
answered Jul 19, 2014 by thanvigandhi_1
 

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