logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

In which region the curve $y^{2}(a+x)=x^{2}(3a-x)$ does not lie?

\[\begin{array}{1 1}(1)\;x>0&(2)\;0< x < 3a \\(3)\;x<-a\; \text{and}\; x>3a & (4)-a \end{array} \]

Can you answer this question?
 
 

1 Answer

0 votes
$y^2 =\large\frac{x^2 (3a-x)}{a+x}$
When $ x < -a$
$x+a < 0$
and $3a-x > 0$
$y^2$ is -ve
y is imaginary
$\therefore$ the curve does not exist in this region.
When $ x > 3a $
$3a-x > 0$
and $a+x > 0$
$y^2$ is -ve
y is imaginary
$\therefore$ the curve does not exist in this region.
When $ -a < x <3a$
Say $x=0$
$x+a=0+a >0 $ and $3a-x=3a-0=3a > 0$
$y^3$ is positive
y is real
The curve exist in these region
Hence 3 is the correct answer.
answered May 20, 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
...