Question

Let $f_1, f_2$ and $f_3$ be three curves satisfying the differential equation $\left(1-y^2\right) d x=2 x y d y$ and pass through $(0,2)$. If $f_3$ cuts the curves $f_1$ and $f_2$ at $A$ and $B$ respectively and one of the curve is passing through $C (5,-1)$, then find the area of $\triangle ABC$.

Answer 1

$\Theta$ Given differential equation can be written as
$ \int \frac{d x}{x}=\int \frac{2 y}{1-y^2} d y=\int \frac{2 y}{y^2-1} d y$
$\Rightarrow \ln x=-\ln \left(y^2-1\right)+\ln c$
$\Rightarrow x\left(y^2-1\right)=c$
$\Theta (0,2) \text { is on it } \Rightarrow c=0$
$\therefore x\left(y^2-1\right)=0 \Rightarrow x=0, y= \pm 1$
Area of $\triangle ABC =\frac{1}{2} \times 2 \times 5=5$