Browse Questions

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 $36x^2+4y^2 = 144$

$\begin{array}{1 1}Eccentricity =\frac{2 \sqrt 2}{3}\quad Latus\; Rectum = \frac{4}{3} \\ Eccentricity =\frac{2 \sqrt 3}{3}\quad Latus \;Rectum = \frac{3}{4} \\Eccentricity =\frac{2 \sqrt 2}{3}\quad Latus \;Rectum = \frac{3}{4} \\ Eccentricity =\frac{3 \sqrt 2}{2}\quad Latus \;Rectum = \frac{4}{3} \end{array}$

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: $36x^2+4y^2 = 144 \rightarrow \large\frac{x^2}{4}$$+\large\frac{y^2}{36}$$=1$
Since $4 \lt 36$, he 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 {4} = 2$ and $a = \sqrt {36} = 6$
$\Rightarrow c = \sqrt {a^2-b^2} = \sqrt{36-4} = \sqrt{32} = 4\sqrt 2$
1) The coordinates of the foci are $(0, 4\sqrt{2})$ and $(0,-4\sqrt{2})$
2) The coordinates of the vertex are $(0,6)$ and $(0,-6)$
3) The Length of the major axis, $2a = 2\times6= 12$
4) The Length of the minor axis, $2b = 2 \times 2= 4$
5) The Eccentricity, $e = \large\frac{c}{a} $$= \large\frac{4\sqrt{2}}{6}$$=\large\frac{2\sqrt 2}{3}$
6) Length of Latus Rectum, $\large\frac{2b^2}{a} $$= \large\frac{2 \times 2^2}{6}$$ = \large\frac{4}{3}$
edited Apr 2, 2014