logo

Ask Questions, Get Answers

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

Find the equation of an ellipse that satisfies the following conditions: Length of minor axis $ = 16$ and Foci $ = (0,\pm6)$

$\begin{array}{1 1}\frac{x^2}{64}+\frac{y^2}{100}=1 \\ \frac{x^2}{16}+\frac{y^2}{36}=1 \\ \frac{x^2}{64}+\frac{y^2}{36}=1 \\ \frac{x^2}{100}+\frac{y^2}{64}=1\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Given an ellipse as follows:
  • http://clay6.com/mpaimg/Toolbar_8.png
  • The equation of the ellipse is $\large\frac{x^2}{b^2}$$+\large\frac{y^2}{a^2}$$=1$
  • Compare the given equation to the general equation of the ellipse to infer $a$ and $b$.
  • $c = \sqrt {a^2 - b^2}$
  • Given minor axis a, and foci c, we can calculate b and substitute back in the equation above.
Length of minor axis $ = 16$ and Foci $ = (0,\pm6) \rightarrow 2b = 16 \rightarrow b = 8$ and $c = 6$
$c^2 = a^2 + b^2 \rightarrow 36 = a^2 + 64 \rightarrow a = \sqrt 100 = 10$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{8^2}$$+\large\frac{y^2}{{10}^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{64}$$+\large\frac{y^2}{100}$$=1$
answered Apr 3, 2014 by balaji.thirumalai
 

Related questions

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