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Find the equation of the ellipse if the centre is $(3 , -4 ), $one of the foci is $(3+\sqrt{3},-4)$ 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}$$=1$
  • http://clay6.com/mpaimg/2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/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)
  • General form of standard ellipses with centre $C$ with major axis $2a$,minor axis $2b$ and axes parallel to the coordinate axes.
  • $\large\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}$$=1$
  • In this case major axis y=k and minor axis $x=h$.
  • $\large\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}$$=1$
  • In this case major axis $x=h$ and minor axis $y=k$.
Step 1:
Centre $C(3,-4)$ and one focus is $F(3+\sqrt 3,-4)$ with $e=\large\frac{\sqrt 3}{2}$
Now $CF=ae=\sqrt 3$
Therefore $a\large\frac{\sqrt 3}{2}$$=\sqrt 3$
$\Rightarrow a=2$
Step 2:
$b=a\sqrt{1-e^2}$
$\;\;=2\sqrt{1-\large\frac{3}{4}}$
$\;\;=1$
Step 3:
The major axis is along CF whose equation is $y=-4$ which is parallel to the $x-axis$
Therefore the equation of the ellipse is $\large\frac{(x-a)^2}{a^2}+\frac{(y-a)^2}{b^2}$$=1$
$\large\frac{(x-3)^2}{4}+$$(y+4)^2=1$
This is the required equation.
answered Jun 13, 2013 by sreemathi.v
edited Jun 13, 2013 by sreemathi.v
 

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