Question

A curve in the first quadrant is such that the area of the triangle formed in the first quadrant by the $x$-axis, a tangent to the curve at any of its point $P$ and radius vector of the point $P$ is $2sq$. units. If the curve passes through $(2,1)$, and the equation of the curve $xy=k$ then $k$ is equal to

Answer 1

Answer: 2

According to question $\frac{1}{2}\left(x-\frac{y}{m}\right)y=2$
$\Rightarrow xy-\frac{y^2}{m}=4$
$\Rightarrow xy-\frac{y^{2}dx}{dy}=4$
$\Rightarrow \frac{dx}{dy}-\frac{1}{y}x=\frac{-4}{y^2}$
$I.F.=e^{-\int \frac{1}{y}dy}=\frac{1}{y}$
$x\frac{1}{y}=-\int \frac{4}{y^3}dy$
$=\frac{4}{2y^2}+C=\frac{2}{y^2}+C$
$x=2;y=1$
$\Rightarrow C=0$
$\Rightarrow xy=2$