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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the vertex and focus of the parabola $\large\frac{x^2}{64}+\frac{y^2}{100}$=1.

$\begin{array}{1 1}(a)\;(0,\pm 10),(0,\pm 6)\\(b)\;(0,\pm 7),(0,\pm 12)\\(c)\;(0,\pm 8),(0,\pm 12)\\(d)\;(0,\pm 6),(0,\pm 14)\end{array}$

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1 Answer

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Given equation of the ellipse is $\large\frac{x^2}{64}+\frac{y^2}{100}$=1.
Comparing the equation with standard equation, $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$=1.
we get $a=8$ and $b=10$.
Since $a < b$, $a^2=b^2(1-e^2)$
$\Rightarrow\:e=\large\frac{3}{5}$
Foci are $S(0,be)$ and $S'(0,-be)$
$S(0,6),\:S'(0,-6)$
$b=10$
$\Rightarrow\:Vertices =(0,b)\: and \:(0,-b)$
Vertices are $ (0,10) \:and\: (0,-10)$
Hence (a) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
edited Mar 24, 2014 by rvidyagovindarajan_1
 

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