The equation of the curve passing through the point $(0,-2)$ given that at any point $(x, y)$ on the curve,…
The equation of the 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 $y$-co-ordinate of the point is equal to the $x$-co-ordinate of the point, is
$y^2+x^2=4$
$y^2-x^2=4$
$2 y^2+x^2=8$
$4 y^2+3 x^2=16$
Solution
$\begin{aligned}
& A / Q \frac{d y}{d x} \cdot y=x \\
& \Rightarrow \int y d y=\int x d x \\
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+c
\end{aligned}$
putting $x=0$ and $y=-2$ we get $c=2$
$\begin{aligned}
& \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+2 \\
& \Rightarrow y^2-x^2=4
\end{aligned}$