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

$\begin{array}{1 1}Eccentricity =\frac{\sqrt 5}{3}\quad Latus\; Rectum = \frac{16}{3} \\ Eccentricity =\frac{\sqrt 20}{3}\quad Latus\; Rectum = \frac{32}{3} \\ Eccentricity =\frac{\sqrt 5}{6}\quad Latus\; Rectum = \frac{16}{3} \\Eccentricity =\frac{\sqrt 20}{3}\quad Latus\; Rectum = \frac{8}{3} \end{array} $

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Toolbox:
  • Given an ellipse as follows:
  • http://clay6.com/mpaimg/Toolbar_7.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: $\large\frac{x^2}{36}$$+\large\frac{y^2}{16}$$=1$
Since $36 \gt 16$, he major 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 a = \sqrt {36} = 6$ and $b = \sqrt {16} = 4$
$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{36-16} = \sqrt{20}$
1) The coordinates of the foci are $(\sqrt{ 20}, 0)$ and $(-\sqrt{ 20}, 0)$
2) The coordinates of the vertex are $(6,0)$ and $(-6,0)$
3) The Length of the major axis, $2a = 2 \times 6 = 12$
4) The Length of the minor axis, $2b = 2 \times 4 = 8$
5) The Eccentricity, $e = \large\frac{c}{a} $$ = \large\frac{\sqrt{20}}{6}$$ = \large\frac{\sqrt 5}{3}$
6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$ = \large\frac{2 \times 4^2}{6} $$ = \large\frac{16}{3}$
answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai
 

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