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Find the equation of the ellipse if the vertices are $(\pm 4 , 0)$and $e=\large\frac{\sqrt{3}}{2}$

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Toolbox:
  • Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
  • $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
  • http://clay6.com/mpaimg/4_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $ (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/4_2_Toolbox1.png
  • Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
  • End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae)$ and $(\pm \large\frac{b^2}{a}$$,-ae)$
  • Directrices $y=\pm \large\frac{a}{e}$.
  • The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
Step 1:
The verices are $(\pm 4,0)$ and $e=\large\frac{\sqrt 3}{2}$
Since the centre is the mid point of the line joining the vertices,the centre is at $(0,0)$.
Also $a=CA=4$
$b=a\sqrt{1-e^2}$
$\Rightarrow 4\sqrt{1-\large\frac{3}{4}}$
$\Rightarrow \large\frac{4}{2}$$=2$
Step 2:
Vertices lie on the major axis is the $x$-axis.
Therefore the equation is $\large\frac{x^2}{16}+\frac{y^2}{4}$$=1$
answered Jun 17, 2013 by sreemathi.v
 

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