Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve,the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Can you answer this question?

1 Answer

0 votes
  • $\int x = \large\frac{x^2}{2}$$ + C$
Step 1:
According to the given information we get the equation as $y\large\frac{dy}{dx}$$ = x $
$ydy = xdx$
Integrating on both sides we get,
$\large\frac{y^2}{2} =\frac{ x^2}{2}$$ + C$
$y^2 - \large\frac{x^2}{2}$$ = C$
Step 2:
Since the curve passes through (0,-2), we substituting for $x$ and $y$ to find the value for C
$(-2)^2 - 0 = 2C$
$C =\large\frac{ 4}{2}$$ = 2$
Substituting this we get
$y^2 - x^2 = 4.$
This is the required solution.
answered Aug 14, 2013 by sreemathi.v

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