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 $\frac{d y}{d x}$.
According to given question,
Product of the slope of tangent with $y$-coordinate $=x$-coordinate
$y \cdot \frac{d y}{d x}=x ......$(i)
On separating the variables, we get
$y d y=x d x$
On integrating both sides, we get $\int y d y=\int x d x$
$\Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+C .....$(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.