Question

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

• A. $y^2+x^2=4$
• B. $x^2-y^2=4$
• C. $y^2-x^2=4$
• D. $x^2+y^2=-4$

Answer 1

Answer: (C)

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 given question,
Product of the slope of tangent with $y$-coordinate $=x$-coordinate
$y\cdot \frac{dy}{dx}=x......$(i)
On separating the variables, we get
$ydy=xdx$
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
$\frac{y^2}{2}=\frac{x^2}{2}+2\Rightarrow x^2-y^2+4=0$
$\Rightarrow y^2-x^2=4$
which is the required equation of the curve.