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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse : $ 4x^2+9y^2=36$

$\begin {array} {1 1} (A)\;Axis : Major \: axis \: along \: x - axis, vertices : ( \pm 3, 0), foci : ( \pm \sqrt 5 , 0), Length \: of \: the \: major \: axis : 6, Length \: of \: the \: minor \: axis : 4, eccentricity \: e = \sqrt 5 / 3, length\: of \: the \: latus \: rectum 8/3 \\ (B)\; Axis : Major \: axis \: along \: x - axis, vertices : ( \pm 3, 0), foci : ( \pm 3 \sqrt 5 , 0), Length \: of \: the \: major \: axis : 6, Length \: of \: the \: minor \: axis : 4, eccentricity \: e = \sqrt 5 / 3, length\: of \: the \: latus \: rectum 8/3 \\ (C)\;Axis : Major \: axis \: along \: y - axis, vertices : ( 0, \pm 3), foci : ( 0, \pm \sqrt 5 ), Length \: of \: the \: major \: axis : 6, Length \: of \: the \: minor \: axis : 4, eccentricity \: e = \sqrt 5 / 3, length\: of \: the \: latus \: rectum 8/3 \\ (D)\;Axis : Major \: axis \: along \: y - axis, vertices : (0, \pm 3), foci : (0, \pm 3\sqrt 5), Length \: of \: the \: major \: axis : 6, Length \: of \: the \: minor \: axis : 4, eccentricity \: e = \sqrt 5 / 3, length\: of \: the \: latus \: rectum 8/3 \end {array}$

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Toolbox:
  • Equation of an ellipse along major axis is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • $c = \sqrt{a^2-b^2}$, where c is the focus of the ellipse.
  • Coordinates of vertices are $( \pm a, 0)$
  • Length of the latus rectum is $\large\frac{2b^2}{a}$
  • Eccentricity $e=\large\frac{c}{a}$
  • Length of the major axis is 2a ; Length of the minor axis is 2b.
Step 1 :
The given equation is $4x^2+9y^2=36$
Dividing throughout by 36 we get,
$\large\frac{x^2}{9}$$+\large\frac{y^2}{4}$$=1$
Comparing this equation with the equation of ellipse.
$ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
It is clear that the major axis is along x - axis and the minor axis is along y - axis.
$ \therefore c = \sqrt{a^2-b^2}$
$ = \sqrt{9-4}$
$ = \sqrt{5}$
Step 2 :
Hence the coordinates of the foci are $( \pm \sqrt 5 , 0)$
The coordinates of the vertices are $( \pm 3, 0)$
Length of the major axis 2a = 6.
Length of the minor axis 2b = 4
eccentricity $e = \large\frac{c}{a}$$ = \large\frac{\sqrt 5 }{3}$
Length of the latus rectum is $ \large\frac{2b^2}{a}$$ = \large\frac{2 \times 4 }{3}$$ = \large\frac{8}{3}$
answered Jul 16, 2014 by thanvigandhi_1
 

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