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,
\[0 - y = \frac{dy}{dx}(X - x)\]
\[\text{ Therefore, }X - x = - y\frac{dx}{dy}\]
\[ \Rightarrow X = x - y\frac{dx}{dy}\]
\[\text{ Therefore, cut off by the tangent on the }x -\text{ axis }= x - y\frac{dx}{dy}\]
\[\text{ Given, }x - y\frac{dx}{dy} = 4y\]
\[\text{ Therefore, }- y\frac{dx}{dy} = 4y - x\]
\[ \Rightarrow \frac{dx}{dy} = \frac{x - 4y}{y}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y}{x - 4y} . . . . . . . . (1)\]
this is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx} \text{ in }(1) \text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{vx}{x - 4vx}\]
\[\text{ Therefore, }v + x\frac{dv}{dx} = \frac{v}{1 - 4v}\]
\[ \Rightarrow \frac{xdv}{dx} = \frac{v}{1 - 4v} - v = \frac{4 v^2}{1 - 4v}\]
\[ \Rightarrow \frac{1 - 4v}{v^2}dv = 4\frac{dx}{x}\]
Integrating on both sides we get,
\[\int\frac{1 - 4v}{v^2}dv = 4\int\frac{dx}{x}\]
\[\text{ Therefore, }\int\frac{dv}{v^2} - 4\int\frac{dv}{v} = 4\int\frac{dx}{x}\]
\[ \Rightarrow - \frac{1}{v} - 4 \log v = 4 \log x + \log c\]
\[ \Rightarrow - \frac{1}{v} = 4 log x + log c + 4 log v\]
\[ \Rightarrow 4 \log(xv) + \log c = - \frac{1}{v}\]
putting the value of v we get
\[4 log(x \times \frac{y}{x}) + log c = - \frac{x}{y}\]
\[ \Rightarrow 4 log(y) + log c = - \frac{x}{y}\]
\[ \Rightarrow \log ( y^4 c) = - \frac{x}{y}\]
\[ \Rightarrow y^4 c = e^{- \frac{x}{y}} \]