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