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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections

Find the equation of an ellipse that satisfies the following conditions: $a = 4$ and Foci = $(\pm3,0)$

$\begin{array}{1 1}\frac{x^2}{16}+\frac{y^2}{7}=1 \\ \frac{x^2}{16}+\frac{y^2}{7}=1 \\ \frac{x^2}{7}+\frac{y^2}{16}=1 \\ \frac{x^2}{7}-\frac{y^2}{16}=1\end{array} $

1 Answer

  • Given an ellipse as follows:
  • The equation of the ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
  • $c= \sqrt {a^2 - b^2}$
  • Given a and foci, we know a and c, so we can calculate b and we can substitute in the above equation and arrive at the equation.
Given Ends of Major axis $\;(\pm3,0)$, Ends of Minor Axis $\;(0, \pm 2)$, $a=4$ and $c = 3$
$c^2 = a^2 - b^2 \rightarrow 9 = 16 - b^2 \rightarrow b = \sqrt{7}$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{4^2}$$+\large\frac{y^2}{\sqrt 7 ^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{16}$$+\large\frac{y^2}{7}$$=1$
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