logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

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$

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

Can you answer this question?
 
 

1 Answer

0 votes
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 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...