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

$\begin{array}{1 1}Eccentricity =\frac{\sqrt 13}{7}\quad Latus\; Rectum = \frac{72}{7} \\ Eccentricity =\frac{\sqrt 13}{6}\quad Latus \;Rectum = \frac{72}{7} \\Eccentricity =\frac{\sqrt 13}{7}\quad Latus \;Rectum = \frac{36}{7} \\ Eccentricity =\frac{\sqrt 13}{6}\quad Latus \;Rectum = \frac{36}{7} \end{array} $

1 Answer

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}{49}$$+\large\frac{y^2}{36 }$$=1$
Since $49 \gt 36$, 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 {49} = 7$ and $b = \sqrt {36} =6$
$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{49-36} = \sqrt{13}$
1) The coordinates of the foci are $(\sqrt{ 13}, 0)$ and $(-\sqrt{ 13}, 0)$
2) The coordinates of the vertex are $(7,0)$ and $(-7,0)$
3) The Length of the major axis, $2a = 2 \times 7= 14$
4) The Length of the minor axis, $2b = 2 \times 3 = 6$
5) The Eccentricity, $e = \large\frac{c}{a} $$ = \large\frac{\sqrt{13}}{7}$
6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$ = \large\frac{2 \times 6^2}{7} $$ = \large\frac{72}{7}$
answered Apr 1, 2014 by balaji.thirumalai
 

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