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

$\begin {array} {1 1} (A)\;Axis : Major \: axis \: is\: along \: the \: x - axis, foci : ( \pm 4\sqrt 2, 0), vertices : ( \pm 6, 0), Length \: of \: major\: axis = 12, length \: of \: the \: minor \: axis : 4, eccentricity \: e = 2\sqrt 2 / 3, length \; of \: the \: latus \: rectum = 4/3 \\ (B)\;Axis : Major \: axis \: is\: along \: the \: y - axis, foci : (0, \pm 4\sqrt 2), vertices : (0, \pm 6), Length \: of \: major\: axis = 12, length \: of \: the \: minor \: axis : 4, eccentricity \: e = 2\sqrt 2 / 3, length \; of \: the \: latus \: rectum = 4/3 \\ (C)\;Axis : Major \: axis \: is\: along \: the \: x - axis, foci : ( \pm2 \sqrt 2, 0), vertices : ( \pm 6,0), Length \: of \: major\: axis = 12, length \: of \: the \: minor \: axis : 4, eccentricity \: e = 4\sqrt 2 / 3, length \; of \: the \: latus \: rectum = 4/3 \\ (D)\;Axis : Major \: axis \: is\: along \: the \: y - axis, foci : (0, \pm 2\sqrt 2), vertices : ( 0, \pm 6), Length \: of \: major\: axis = 12, length \: of \: the \: minor \: axis : 4, eccentricity \: e = 4\sqrt 2 / 3, length \; of \: the \: latus \: rectum = 4/3 \end {array}$

## 1 Answer

Toolbox:
• If $b > a$ then the major axis is y - axis and the minor axis is along x - axis. Hence the equation of the ellipse is $\large\frac{x^2}{b^2}$$+ \large\frac{y^2}{a^2}$$=1$.
• $c = \sqrt{a^2-b^2}$ where c is the focus of the ellipse.
• Coordinates of foci are $( 0, \pm c)$
• Coordinates of vertices are $(0, \pm b)$
• Length of the latus rectum = $\large\frac{2b^2}{a}$
• eccentricity $e = \large\frac{c}{a}$
• Length of the major axis is 2a ; length of the minor axis is 2b.
Step 1 :
The given equation is $36x^2+4y^2=144$
It can be written as
$\large\frac{x^2}{4}$$+ \large\frac{y^2}{36}$$=1$
Comparing this with the equation of the ellipse
$\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$ we get,
$a^2=36$ and $b^2=4$
It is clear that the major axis is along y axis and minor axis is along x - axis.
$c = \sqrt{a^2-b^2}$
$= \sqrt{36-4}$
$= \sqrt{32}$
$4\sqrt 2$
The coordinates of foci are $(0, \pm 4\sqrt 2 )$
the coordinates of vertices are $(0, \pm 6)$
eccentricity $= \large\frac{c}{a}$$= \large\frac{4\sqrt 2}{6}$$ = \large\frac{2\sqrt 2 }{3}$
Length of the major axis 2a : 12
Length of the minor axis 2b : 4
Length of latus rectum : $\large\frac{2b^2}{a}$$= \large\frac{2 \times 4 }{5}$$ = \large\frac{4}{3}$
answered Jul 16, 2014

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