Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
  • Given an ellipse as follows:
  • http://clay6.com/mpaimg/Toolbar_7.png
  • 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$
answered Apr 3, 2014 by balaji.thirumalai
edited Apr 3, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App