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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 : $ \large\frac{x^2}{100}$$+ \large\frac{y^2}{400}$$=1$

$\begin {array} {1 1} (A)\;Axis : Major \: axis \: is\: along \: the \: x - axis, foci : ( \pm 10\sqrt 3, 0), vertices : ( \pm 20, 0), Length \: of \: major\: axis = 40, length \: of \: the \: minor \: axis : 20, eccentricity \: e = \sqrt 3 / 2, length \; of \: the \: latus \: rectum = 10 \\ (B)\; Axis : Major \: axis \: is\: along \: the \: x - axis, foci : ( \pm \sqrt 3, 0), vertices : ( \pm 40, 0), Length \: of \: major\: axis = 40, length \: of \: the \: minor \: axis : 20, eccentricity \: e = \sqrt 3 / 2, length \; of \: the \: latus \: rectum = 10 \\ (C)\;Axis : Major \: axis \: is\: along \: the \: y - axis, foci : ( 0, \pm \sqrt 3), vertices : (0, \pm 40), Length \: of \: major\: axis = 40, length \: of \: the \: minor \: axis : 20, eccentricity \: e = \sqrt 3 / 2, length \; of \: the \: latus \: rectum = 10 \\ (D)\;Axis : Major \: axis \: is\: along \: the \: y - axis, foci : (0, \pm 10\sqrt 3), vertices : ( 0, \pm 20), Length \: of \: major\: axis = 40, length \: of \: the \: minor \: axis : 20, eccentricity \: e = \sqrt 3 / 2, length \; of \: the \: latus \: rectum = 10 \end {array}$

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  • If $b > a $ then the major axis is y - axis and the minor axis is along x - axis. Hence the equation of the ellipse is $ \large\frac{x^2}{b^2}$$ + \large\frac{y^2}{a^2}$$=1$.
  • $c = \sqrt{a^2-b^2}$ where c is the focus of the ellipse.
  • Coordinates of foci are $( 0, \pm c)$
  • Coordinates of vertices are $(0, \pm b)$
  • Length of the latus rectum = $\large\frac{2b^2}{a}$
  • 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 $ \large\frac{x^2}{100}$$+ \large\frac{y^2}{400}$$=1$
Comparing this with the equation of the ellipse $ \large\frac{x^2}{b^2}$$+\large\frac{x^2}{a^2}$$=1$ we get,
$a^2=400$ and $ b^2=100$
$ \therefore $ The major axis is along the y - axis and the minor axis is along the x - axis.
Hence $c = \sqrt{a^2-b^2}$
$ = \sqrt{400-100}$
$= \sqrt{300}$
$ = 10\sqrt 3 $
Step 2 :
$ \therefore $ The coordinates of the foci are $(0, \pm 10\sqrt 3 )$
The coordinates of the vertices are $(0, \pm 20)$
Length of the major axis : 2a = 40
Length of the minor axis : 2b = 20
eccentricity $e = \large\frac{c}{a}$$ = \large\frac{10\sqrt 3}{20}$$ = \large\frac{\sqrt 3 }{2}$
Length of the latus rectum = $ \large\frac{2b^2}{a}$$ = \large\frac{2 \times 100}{20}$$ = 10$
answered Jul 16, 2014 by thanvigandhi_1
 

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