Let $y=f(x)$ be a curve passing through $(1,1)$ such that the triangle formed by the coordinate axes

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 $1^{	ext {st }}$ quadrant and has area 2. The possible curves are

Professor PC Thomas Classes · Answer

Equation of the tangent at $(x, y)$ to the curve is $Y-y=M(X-x)$ where $M$ is the slope of the tangent $=f^{\prime}(x)$.\ This meets the $x$-axis at $P\left(x-\frac{y}{M}, 0ight)$, and $y$-axis at $Q(0, y-x M)$\ Area of $	riangle O P Q=2 \Rightarrow \frac{-(y-x M)^{2}}{2 M}=2$\ $\Rightarrow(y-M x)^{2}+4 M=0$\ $\Rightarrow \quad M<0$ as $(y-M x)^{2}>0$\ $	herefore y-M x=2 \sqrt{-M} \Rightarrow y=M x+2 \sqrt{-M}$\ Differentiating w.r.t. $x$,\ $\frac{d y}{d x}=M+x \frac{d M}{d x}+2 \frac{1}{2(\sqrt{-M})}\left(-\frac{d M}{d x}ight)$\ $	herefore \frac{d y}{d x}=M+\left[x-(-M)^{-1 / 2}ight] \frac{d M}{d x}$\ If $\frac{d M}{d x}=0$ then $M=c$, a constant\ Using this in (1) $y=c x+2 \sqrt{-c}$\ This passes through $(1,1) \Rightarrow c=-1$\ $	herefore \quad y=-x+2$ or $x+y=2$\ If $x=(-M)^{-1 / 2}$ then $x^{2}=-\frac{d x}{d y}$\ Solving $x y=1$\ $	herefore \quad$ Solutions are (i.e. the two required curves are)\ $x+y=2$ and $x y=1$\