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7. 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.

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

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Given

The equation of the ellipse

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

$\Rightarrow \frac{36}{144}x^2 + \frac{4}{144}y^2 =1$

$\Rightarrow \frac{1}{4}x^2 + \frac{1}{36}y^2 =1$

$\frac{x^2}{2^2} + \frac{y^2}{6^2} = 1$

As we can see from the equation, the major axis is along Y-axis and the minor axis is along X-axis.

On comparing the given equation with the standard equation of such ellipse, which is

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$

We get

$a=6$ and $b=2$ .

So,

$c=\sqrt{a^2-b^2}=\sqrt{6^2-2^2}$

$c=\sqrt{32}=4\sqrt{2}$

Hence,

Coordinates of the foci:

$(0,c)\:and\:(0,-c)=(0,4\sqrt{2})\:and\:(0,-4\sqrt{2})$

The vertices:

$(0,a)\:and\:(0,-a)=(0,6)\:and\:(0,-6)$

The length of the major axis:

$2a=2(6)=12$

The length of minor axis:

$2b=2(2)=4$

The eccentricity :

$e=\frac{c}{a}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}$

The length of the latus rectum:

$\frac{2b^2}{a}=\frac{2(2)^2}{6}=\frac{8}{6}=\frac{4}{3}$

Posted by

Pankaj Sanodiya

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