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

$\begin {array} {1 1} (A)\;Foci : (0, \pm \sqrt{21} ), vertices : (0, \pm 5), length \: of \: the \: major \: axis = 10, length \: of \: the \: minor\: axis = 4, eccentricity = \large\frac{\sqrt{21}}{5}, length \: of \: the \: latus \: rectum = \large\frac{8}{5}, axis \: major \: axis \: along \: y - axis \\ (B)\;Foci : ( \pm \sqrt{21} , 0), vertices : ( \pm 5, 0), length \: of \: the \: major \: axis = 10, length \: of \: the \: minor\: axis = 4, eccentricity = \large\frac{\sqrt{21}}{5}, length \: of \: the \: latus \: rectum = \large\frac{8}{5}, axis \: major \: axis \: along \: x - axis \\ (C)\;Foci : (0, \pm 2\sqrt{21} ), vertices : (0, \pm 5), length \: of \: the \: major \: axis = 10, length \: of \: the \: minor\: axis = 4, eccentricity =2 \large\frac{\sqrt{21}}{5}, length \: of \: the \: latus \: rectum = \large\frac{16}{5}, axis \: major \: axis \: along \: y - axis \\ (D)\;Foci : ( \pm 2\sqrt{21}, 0 ), vertices : ( \pm 5, 0), length \: of \: the \: major \: axis = 10, length \: of \: the \: minor\: axis = 4, eccentricity = \large\frac{2\sqrt{21}}{5}, length \: of \: the \: latus \: rectum = \large\frac{16}{5}, axis \: major \: axis \: along \: x - axis \end {array}$

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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.
The given equation is $ \large\frac{x^2}{4}$$+\large\frac{y^2}{25}$$=1$
Since $b > a$ , the major axis is along the y - axis, while the minor axis is along the x - axis.
Comparing the equation with
$\large\frac{x^2}{b^2}$$+ \large\frac{y^2}{a^2}$$=1$
We get, b = 2 and a = 5
$ \therefore c = \sqrt{a^2-b^2}$
$ = \sqrt{25-4}$
$ = \sqrt{21}$
Hence the foci are $(0, \sqrt{21})$ and $(0, -\sqrt{21} )$ The coordinates of the vertices are (0,5) and (0,-5).
Length of the major axis = $2a=2 \times 5 = 10$
Length of the minor axis = $2b=2 \times 2 = 4$
Eccentricity $ e = \large\frac{c}{a}$$ = \large\frac{\sqrt{21}}{5}$
Length of the latus rectum = $ \large\frac{2b^2}{a}$$ = \large\frac{2 \times 4 }{5} $$ = \large\frac{8}{5}$
answered Jul 15, 2014 by thanvigandhi_1
 

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