Question

Which of the following equations represents the curve for which the intercept cut-off by any tangent on y-axis is proportional to the square of the ordinate of the point of tangency?

• A. $x+y=cxy$
• B. $\frac{1}{x}+\frac{1}{y}=c$
• C. $\frac{A}{x}+\frac{B}{y}=xy$
• D. $\frac{A}{x}+\frac{B}{y}=1$

Answer 1

Answer: (D)

The equation of tangent at any point P(x,y) is
$Y-y\frac{dy}{dx}\left(X-x\right)$

For Y intercept put $X=0$
$\Rightarrow Y=y-x\frac{dy}{dx}$
Given $Y\propto y^2\Rightarrow y-x\frac{dy}{dx}=k{y}^2$
[k proportionality constant]
$\Rightarrow \frac{dy}{dx}-\frac{1}{x}y=-\frac{k{y}^2}{x}\Rightarrow y^{-2}\frac{dy}{dx}-\frac{1}{x}{y}^{-1}=-\frac{k}{x}$
Put $y^{-1}=z\Rightarrow -y^{-2}\frac{dy}{dx}=\frac{dz}{dx}$.
Then $-\frac{dz}{dx}-\frac{1}{x}z=-\frac{k}{x}\Rightarrow \frac{dz}{dx}+\left(\frac{1}{x}\right)z=\frac{k}{x}$
$I.F.=e^{\int \frac{1}{x}dx}=e^{logx}=x$
The solution is $z\left(x\right)=\int \frac{k}{x}\left(x\right)dx+c$
$y^{-1}x=kx+c\Rightarrow \frac{x}{y}=kx+c\Rightarrow \frac{1}{ky}=1+\frac{c}{kx}$
$\Rightarrow \frac{\left(\frac{1}{k}\right)}{y}+\frac{\left(-\frac{c}{k}\right)}{x}=1\Rightarrow \frac{A}{x}+\frac{B}{y}=1$