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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 : $ 16x^2 +y^2= 16$

$\begin {array} {1 1} (A)\;Axis : Major \: axis \: along \: x - axis, foci : (\pm 5\sqrt 3, 0), coordinates \: of vertices : ( \pm 4, 0), Length\: of\: the\: major\: axis : 8, length \: of \: the \: minor \: axis : 2, eccentricity \: e = \sqrt{15}/4, length \: of \: the \: latus \: rectum = 1/2 \\ (B)\;Axis : Major \: axis \: along \: x - axis, foci : (\pm \sqrt 3, 0), coordinates \: of vertices : ( \pm 4, 0), Length\: of\: the\: major\: axis : 8, length \: of \: the \: minor \: axis : 2, eccentricity \: e = \sqrt{15}, length \: of \: the \: latus \: rectum = 1/2 \\ (C)\;Axis : Major \: axis \: along \: y - axis, foci : (0, \pm \sqrt{15}), coordinates \: of vertices : ( 0,\pm 4), Length\: of\: the\: major\: axis : 8, length \: of \: the \: minor \: axis : 2, eccentricity \: e = \sqrt{15}/4, length \: of \: the \: latus \: rectum = 1/2 \\ (D)\;Axis : Major \: axis \: along \: y - axis, foci : (0, \pm \sqrt 3, 0), coordinates \: of vertices : ( \pm 4, 0), Length\: of\: the\: major\: axis : 8, length \: of \: the \: minor \: axis : 2, eccentricity \: e = \sqrt{15}/4, length \: of \: the \: latus \: rectum = 1/2 \end {array}$

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1 Answer

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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 $16x^2+y^2=16$.
Dividing through out by 16 we get,
$\large\frac{x^2}{1}$$+\large\frac{y^2}{16}$$=1$
Comparing this equation with the equation of ellipse
$ \large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
$a^2=16$ and $ b^2=1$
It is clear that the major axis is along y - axis.
$ c = \sqrt{a^2-b^2}$
$ = \sqrt{16-1}$
$ = \sqrt{15}$
Step 2 :
$ \therefore $ The coordinates of foci are $(0, \pm \sqrt{15})$
The coordinates of vertices are $(0, \pm 4)$
Length of major axis 2a = 8
Length of minor axis 2b = 2
Eccentricity $ e = \large\frac{c}{a} $$=\large\frac{\sqrt{15}}{4}$
Length of thelatus rectum = $ \large\frac{2b^2}{a}$$ = \large\frac{2 \times 1}{2}$$ = \large\frac{1}{2}$
answered Jul 16, 2014 by thanvigandhi_1
 

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