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 $2\, sq$. units. If the curve passes through $(2,1)$, and the equation of the curve $x y=k$ then $k$ is equal to

Answer 1

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