Question

Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

Answer 1

Let the given curve be y = f(x). Suppose P(x,y) be a point on the curve. Equation of the tangent to the curve at P is
$Y-y=\frac{dy}{dx}(X-x)$ , where (X, Y) is the arbitrary point on the tangent.
Putting Y=0 we get,
$$\begin{gathered} 0-y=\frac{dy}{dx}(X-x) \\ \mathrm{Therefore},\mathrm{X}-x\mathit{=}\mathit{-}y\frac{dx}{dy} \\ \Rightarrow \mathrm{X}=x\mathit{-}y\frac{dx}{dy} \\ \mathrm{Therefore},\mathrm{cut}\mathrm{off}\mathrm{by}\mathrm{the}\mathrm{tangent}\mathrm{on}\mathrm{the}\mathrm{x}-\mathrm{axis}\mathit{}\mathit{=}\mathit{}x\mathit{-}y\frac{dx}{dy} \\ Given\mathit{,}\mathit{}x\mathit{-}y\frac{dx}{dy}\mathit{=}\mathit{4}y \\ Therefore\mathit{,}\mathit{}\mathit{-}y\frac{dx}{dy}\mathit{=}\mathit{4}y\mathit{}\mathit{-}\mathit{}x \\ \mathit{\Rightarrow }\frac{dx}{dy}\mathit{=}\frac{x\mathit{-}\mathit{4}y}{y} \\ \mathit{\Rightarrow }\frac{dy}{dx}\mathit{=}\frac{y}{x\mathit{-}\mathit{4}y}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\left(1\right) \\ \mathrm{this}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \end{gathered}$$
$$\begin{gathered} \mathrm{Putting}y=vxand\frac{dy}{dx}=v+x\frac{dv}{dx}in\left(1\right)\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{vx}{x-4vx} \\ Therefore,v+x\frac{dv}{dx}=\frac{v}{1-4v} \\ \Rightarrow \frac{xdv}{dx}=\frac{v}{1-4v}-v=\frac{4v^2}{1-4v} \\ \Rightarrow \frac{1-4v}{v^2}dv=4\frac{dx}{x} \end{gathered}$$
Integrating on both sides we get,
$$\begin{gathered} \int \frac{1-4v}{v^2}dv=4\int \frac{dx}{x} \\ Therefore,\int \frac{dv}{v^2}-4\int \frac{dv}{v}=4\int \frac{dx}{x} \\ \Rightarrow -\frac{1}{v}-4\log v=4\log x+\log c \\ \Rightarrow -\frac{1}{v}=4logx+logc+4logv \\ \Rightarrow 4\log \left(xv\right)+\log c=-\frac{1}{v} \\ puttingthevalueofvweget \\ 4log(x\times \frac{y}{x})+logc=-\frac{x}{y} \\ \Rightarrow 4log\left(y\right)+logc=-\frac{x}{y} \\ \Rightarrow \log \left(y^{4}c\right)=-\frac{x}{y} \\ \Rightarrow y^{4}c=e^{-\frac{x}{y}} \end{gathered}$$