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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}{25}$$+\large\frac{y^2}{100}$$=1$

$\begin {array} {1 1} (A)\;Axis : Major \: axis \: along \: x - axis, Foci : ( \pm 5\sqrt 3, 0 ), Vertices ( \pm 10, 0), length \: of \: the\: major\: axis : 20 , length \: of\: the\: minor\: axis : 10, Eccentricity \: e = \sqrt 3/ 2, length \: of \: the \: latus \: rectum : 5 \\ (B)\;Axis : Major \: axis \: along \: y - axis, Foci : ( 0, \pm 5\sqrt 3 ), Vertices (0, \pm 10), length \: of \: the\: major\: axis : 20 , length \: of\: the\: minor\: axis : 10, Eccentricity \: e = \sqrt 3/ 2, length \: of \: the \: latus \: rectum : 5 \\ (C)\;Axis : Major \: axis \: along \: y - axis, Foci : (0, \pm \sqrt 3 ), Vertices (0, \pm 5), length \: of \: the\: major\: axis : 5 , length \: of\: the\: minor\: axis : 10, Eccentricity \: e = \sqrt 3/ 2, length \: of \: the \: latus \: rectum : 5 \\ (D)\;Axis : Major \: axis \: along \: x - axis, Foci : ( \pm \sqrt 3 , 0), Vertices ( \pm 5, 0), length \: of \: the\: major\: axis : 10 , length \: of\: the\: minor\: axis : 5, Eccentricity \: e = 2\sqrt 3, length \: of \: the \: latus \: rectum : 5 \end {array}$

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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}{25}$$+\large\frac{y^2}{100}$$=1$
Comparing this with equation of the ellipse $ \large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
We get, $a^2=100$ and $b^2=25$
Hence the major axis is along y - axis.
$ \therefore c = \sqrt{a^2-b^2} = \sqrt{100-25} = 75 = 5 \sqrt 3$
Step : 2
The coordinates of foci are $(0, \pm 5 \sqrt 3)$
The coordinates of vertices are $(0, \pm 10)$
Length of major axis $= 2a = 20$
Length of minor axis $=2b = 10$
Eccentricity $e= \large\frac{c}{a} $$ = \large\frac{5\sqrt 3}{10} = \large\frac{\sqrt 3 }{2}$
Length of latus rectum $ = \large\frac{2b^2}{a^2}$$ = \large\frac{2 \times 25}{10}$$=5$
answered Jul 15, 2014 by thanvigandhi_1
 

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