Find the equation of a curve a segment of which, cut off on the $x$-axis by a tangent line, is equal to the square of the ordinate of the point of tangency.
For a generic point $P(x_o,y_o)$ the tangent equation will be $y-y_o = \frac{dy}{dx}(x-x_o)$. Hence the $x$ intercept will be $x_{int} = -y_o \frac{dx}{dy} + x_o$. Hence the differential equation that must be solved is:
$$x^2 = -y\frac{dx}{dy} + x$$
However the correct answer is $y^2+x+ay=0$. The solution to this differential is a separable one of the result $-\frac{e^{c1}x}{1-e^{c_1}x}$. The cutt of the x-axis, I am interpreting as the distance between the origin and the x intercept. Is this correct?
Best Answer
It says x-intercept is equal to the square of the ordinate of the point of tangency. Ordinate is $y$, you took abscissa by mistake.
So the differential equation should be,
$ \displaystyle y^2 = x_{int} = -y \ \frac{dx}{dy} + x \implies y \ dx - x \ dy + y^2 \ dy = 0 $
Dividing by $y^2$, $ \displaystyle \frac{y \ dx - x \ dy}{y^2} + dy = 0$
$ \displaystyle d\left(\frac{x}{y}\right) + dy = 0$
Integrating, $\displaystyle \frac{x}{y} + y + a = 0 \implies y^2 + ay + x = 0, $ where $a$ is a constant.