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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 the 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}Eccentricity =\frac{\sqrt 15}{4}\quad Latus\; Rectum = \frac{1}{2} \\ Eccentricity =\frac{\sqrt 15}{2}\quad Latus \;Rectum = \frac{1}{2} \\Eccentricity =\frac{\sqrt 15}{4}\quad Latus \;Rectum = \frac{1}{4} \\ Eccentricity =\frac{\sqrt 15}{2}\quad Latus \;Rectum = \frac{1}{4} \end{array} $

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Toolbox:
  • Given an ellipse as follows:
  • http://clay6.com/mpaimg/Toolbar_8.png
  • The equation of the ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • Compare the given equation to the general equation of the ellipse to infer $a$ and $b$.
  • $C = \sqrt {a^2 - b^2}$
  • Coordinates of foci are $(c,0)$ and $(-c,0)$
  • The coordinates of the Vertex are $(a,0)$ and $(-a,0)$
  • Length of major axis = $2a$, Length of minor axis = $2b$
  • Eccentricity $e = \large\frac{c}{a}$
  • Length of Latus Rectum = $\large\frac{2b^2}{a}$
Given: $16x^2+y^2=16 \rightarrow \large\frac{x^2}{1}$$+\large\frac{y^2}{16}$$=1$
Since $1 \lt 16$, the minor axis is along the x-axis.
On comparing the equation with $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1 \rightarrow b = \sqrt {1} = 1$ and $a = \sqrt {16} = 4$
$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{16-1} = \sqrt{15}$
1) The coordinates of the foci are $(0, \sqrt{15})$ and $(0,-\sqrt{15})$
2) The coordinates of the vertex are $(0,4)$ and $(0,-4)$
3) The Length of the major axis, $2a = 2\times4= 8$
4) The Length of the minor axis, $2b = 2 \times 1= 2$
5) The Eccentricity, $e = \large\frac{c}{a} $$ = \large\frac{\sqrt{15}}{4}$
6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$ = \large\frac{2 \times 1^2}{4} $$ = \large\frac{1}{2}$
answered Apr 2, 2014 by balaji.thirumalai
 

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