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Find the equation of the ellipse if the centre is$ (3 , -1 ),$ one of the foci is $(6 , -1)$ and passing through the point $(8 , -1)$.

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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}$$=1$
  • http://clay6.com/mpaimg/2_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/2_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:
$C(3,-1)$ is the centre,$F(6,-1)$ is one focus and the ellipse passes through $P(8,-1).$
$CF$ is the major axis (i.e)$y=-1$ (parallel to the $x$-axis)
Now the point $P$ lies on $y=-1$.
Therefore it is one of the vertices.
Step 2:
$CF=ae$
$LP=a$
Therefore $ae=3$
$a=5$
$e=\large\frac{3}{5}$
Step 3:
$b=5\sqrt{1-\large\frac{9}{25}}$
$\Rightarrow 5.\large\frac{4}{5}$$=4$
Step 4:
The equation of the ellipse is $\large\frac{(x-3)^2}{25}+\large\frac{(y+1)^2}{16}$$=1$
answered Jun 17, 2013 by sreemathi.v
 

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