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
$Y-y=\frac{d y}{d x}(X-x)$

whose, $x$-intercept $\left(x-y \cdot \frac{d x}{d y}, 0\right)$
$y$-intercept $\left(0, y-x \frac{d y}{d x}\right)$
Given, $ \triangle O P Q=2$
$\Rightarrow \frac{1}{2} \cdot\left(x-y \frac{d x}{d y}\right)\left(y-x \frac{d y}{d x}\right)=2$
$\Rightarrow \left(x-y \frac{1}{p}\right)(y-x p)=4$, where $ p=\frac{d y}{d x}$
$\Rightarrow p^{2} x^{2}-2 p x y+4 p+y^{2}=0$
$\Rightarrow (y-p x)^{2}+4 p=0$
$\therefore y-p x=2 \sqrt{-p}$
$\Rightarrow y=p x+2 \sqrt{-p}$
$p=p+\frac{d p}{d x} \cdot x+2 \cdot\left(\frac{1}{2}\right)(-p)^{-1 / 2} \cdot(-1) \frac{d p}{d x}$
$\Rightarrow \frac{d p}{d x}\left\{x-(-p)^{-1 / 2}\right\}=0$
$\Rightarrow \frac{d p}{d x}=0 $ or $ x=(-p)^{-1 / 2}$
On putting this value in Eq. (i), we get $y=c x+2 \sqrt{-c}$
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)
$y=\frac{-x}{x^{2}}+2 \cdot \frac{1}{x} \Rightarrow x y=1$
Thus, the two curves are $x y=1$ and $x+y=2$.