Question

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.

Answer 1

Let x and y be the x-coordinate and y-coordinate of the curve respectively.
We know that the slope of a tangent to the curve in the coordinate axis is given by the relation $\frac{dy}{dx}$
According to the given question, we get
Product of the slope of tangent with y-coordinate =x=coordinate
$y.\frac{dy}{dx}=x$ ...(i)
On separating the variables, we get
y dy=x dx
On integrating both sides, we get $\int ydy=\int xdx\Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+C$ ...(ii)
Now, the curve passes through the point(0,-2), therefore
$\frac{(-2{)}^2}{2}=0+C\Rightarrow C=\frac{4}{2}=2$
On substituting this value of C in Eq. (ii), we get
$\frac{y^2}{2}=\frac{x^2}{2}+2\Rightarrow x^2-y^2+4=0$
which is the required equation of the curve.