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Find the equation of the ellipse if the centre at the origin, the major axis is along $x$-axies,$e=\large\frac{2}{3}$ and passes through the point $(2 , \large\frac{-5}{3})$

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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}$$=1$
  • http://clay6.com/mpaimg/1_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $ (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/1_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 centre is at $(0,0)$,major axis along $x$-axis,$e=\large\frac{2}{3}$
The equation is of the form $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$-----(1)
Where $b^2=a^2(1-e^2)$
$\Rightarrow a^2\big(1-\large\frac{4}{9}\big)=\large\frac{5a^2}{9}$
Substitute this in eq(1) we get
$\large\frac{x^2}{a^2}+\frac{9y^2}{5a^2}$$=1$
Step 2:
The point $(2,-\large\frac{5}{3}\big)$ lies on the ellipse .
Substituting the coordinates we have
$\large\frac{4}{a^2}+\large\frac{9\times 25/7}{5a^2}$$=1$
$\large\frac{1}{a^2}$$[4+5]=1$
$a^2=9$
$a= 3$
$b=\large\frac{\sqrt 5\times 3}{3}$$\Rightarrow \sqrt 5$
The equation of the ellipse is
$\large\frac{x^2}{9}+\frac{y^2}{5}$$=1$
This is the required equation.
answered Jun 13, 2013 by sreemathi.v
edited Jun 13, 2013 by sreemathi.v
 

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