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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}{36}$$+\large\frac{y^2}{16}$$=1$

$\begin {array} {1 1} (A)\;Axis \: :\: major \: axis \: along \: x - axis, focus \: :\: ( \pm 2\sqrt 5, 0), vertices \: : \: ( \pm 6, 0), eccentricity : \sqrt 5 /3, length \: of \: the \: lactus \: rectum = \large\frac{16}{3} \\ (B)\;Axis \: :\: major \: axis \: along \: y - axis, focus \: :\: (0, \pm 2\sqrt 5), vertices \: : \: (0, \pm 6), eccentricity : \sqrt 5 /3, length \: of \: the \: lactus \: rectum = \large\frac{16}{3} \\ (C)\;Axis \: :\: major \: axis \: along \: x - axis, focus \: :\: ( \pm \sqrt 5, 0), vertices \: : \: ( \pm 6, 0), eccentricity :3/ \sqrt 5 , length \: of \: the \: lactus \: rectum = \large\frac{16}{3} \\ (D)\;Axis \: :\: major \: axis \: along \: y - axis, focus \: :\: (0, \pm \sqrt 5), vertices \: : \: ( \pm 6, 0), eccentricity :3/ \sqrt 5 , length \: of \: the \: lactus \: rectum = \large\frac{16}{3} \end {array}$

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  • 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.
Given equation is $\large\frac{x^2}{36}$$+\large\frac{y^2}{16}$$=1$
Since $36 > 16$
$ \Rightarrow a^2 > b^2$ or a > b
The major axis is along x - axis and the minor axis is along y - axis.
Let us compare this equation with the general equation.
$\large\frac{x^2}{a^2}$$+ \large\frac{y^2}{b^2}$$=1$
This implies $a^2=36 $ and $ b^2=16$
$ \therefore c = \sqrt{a^2-b^2}$
= $ \sqrt{36-16}$
$ = \sqrt{20} = 2 \sqrt 5 $
$ \therefore $ The coordinates of foci are $(2\sqrt 5 , 0)$ and $(-2\sqrt 5 , 0)$
The coordinates of the vertices are (6, 0) and (-6, 0)
length of the major axis is $2a = 2 \times 6 = 12$
length of the minor axis is $2b = 2 \times 4 = 8$
Eccentricity $e = \large\frac{c}{a}$$ = \large\frac{2\sqrt 5 }{6}$$ = \large\frac{\sqrt 5}{3}$
length of the latus rectum = $ \large\frac{2b^2}{a}$$ = \large\frac{2 \times 16}{6}$
$ = \large\frac{16}{3}$
answered Jul 15, 2014 by thanvigandhi_1
 

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