Question

Let $y=f(x)$ be a curve passing through $(1,1)$ such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area $2$ unit. Form the differential equation and determine all such possible curves.

Answer 1

Equation of tangent to the curve $y=f(x)$ at point $A(x,y)$ is
$Y-y=\frac{dy}{dx}(X-x)$

whose, $x$-intercept $\left(x-y\cdot \frac{dx}{dy},0\right)$
$y$-intercept $\left(0,y-x\frac{dy}{dx}\right)$
Given, $△OPQ=2$
$\Rightarrow \frac{1}{2}\cdot \left(x-y\frac{dx}{dy}\right)\left(y-x\frac{dy}{dx}\right)=2$
$\Rightarrow \left(x-y\frac{1}{p}\right)(y-xp)=4$, where $p=\frac{dy}{dx}$
$\Rightarrow p^2{x}^2-2pxy+4p+y^2=0$
$\Rightarrow (y-px{)}^2+4p=0$
$\therefore y-px=2\sqrt{-p}$
$\Rightarrow y=px+2\sqrt{-p}$
$p=p+\frac{dp}{dx}\cdot x+2\cdot \left(\frac{1}{2}\right)(-p{)}^{-1/2}\cdot (-1)\frac{dp}{dx}$
$\Rightarrow \frac{dp}{dx}\left\{x-(-p{)}^{-1/2}\right\}=0$
$\Rightarrow \frac{dp}{dx}=0$ or $x=(-p{)}^{-1/2}$
On putting this value in Eq. (i), we get $y=cx+2\sqrt{-c}$
This curve passes through $(1,1)$.
$\Rightarrow 1=c+2\sqrt{-c}\Rightarrow c=-1$
$\therefore y=-x+2\Rightarrow x+y=2$
Again, if $x=(-p{)}^{-1/2}$
$\Rightarrow -p=\frac{1}{x^2}$ putting in Eq. (i)
$y=\frac{-x}{x^2}+2\cdot \frac{1}{x}\Rightarrow xy=1$
Thus, the two curves are $xy=1$ and $x+y=2$.