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 \(\frac{dy}{dx}\)
According to the given information,we get:
\(y.\frac{dy}{dx}=x\)
\(⇒y\, dy=x\, dx\)
Integrating both sides,we get:
\(∫y\, dy=∫x\, dx\)
\(⇒\frac{y^2}{2}=\frac{x^2}{2}+C\)
\(⇒y^2-x^2=2C...(1)\)
Now,the curve passes through point(0,-2).
\(∴(-2)^2-0^2=2C\)
\(⇒2C=4\)
Substituting \( 2C=4\) in equation(1),we get:
\(y^2-x^2=4\)
This is the required equation of the curve.