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Find the eccentricity, centre, foci, vertices of the following ellipses and draw the diagram: $16x^{2}+25y^{2}=400$

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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}$$=1$
  • http://clay6.com/mpaimg/13_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/13_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:
$16x^2+25y^2=400$
The above equation is divided by 400 we get
$\large\frac{x^2}{25}+\frac{y^2}{16}$$=1$
$a^2=25,b^2=16$
$\Rightarrow a=5,b=4$
The major axis is the $x$-axis.
Step 2:
Eccentricity :$e=\sqrt{1-\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1-\large\frac{16}{25}}$
$\Rightarrow \sqrt{\large\frac{9}{25}}$
$\Rightarrow \large\frac{3}{5}$
Step 3:
Centre : The origin (0,0).
Step 4:
Foci $(\pm ae,0)$.
$ae=5\times \large\frac{3}{5}$$=3$
The foci are $(\pm 3,0)$
Step 5:
Vertices : $(\pm a,0)$
$a=5$
Therefore the vertices are $(\pm 5,0)$
answered Jun 18, 2013 by sreemathi.v
edited Jun 21, 2013 by sreemathi.v
 

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