logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Equation of the director circle of the ellipse $x^2+2y^2+2x-12y+15=0$ is

$\begin{array}{1 1}(A)\;x^2+y^2+2x-6y+4=0 \\(B)\;x^2+y^2+2x-12y+4=0 \\(C)\;x^2+y^2+2x-6y-4=0 \\(D)\;x^2+y^2+2x-12y-4=0 \end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
$x^2+2y^2+2x-12y+15=0$
=> $ \large\frac{(x+1)^2}{4}+\frac{(y-3)^2}{2}$$=1$
$\therefore $ Equation of its director circle is
$(x+1)^2+(y-3)^2=4+2$
=> $x^2+y^2+2x-6y+4=0$
Hence A is the correct answer.
answered Apr 10, 2014 by meena.p
 

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
...