## Related questions with answers

Find a family of oblique trajectories that intersect the family of curves $x+$ $y=c x^{2}$ at angle $a$ such that $\tan a=2$.

Solution

VerifiedDifferentiate the equation of the family $x + y = c \ x^2$ with respect to x. Then we have

$\begin{gather*} 1 + \dfrac{dy}{dx} = 2c \ x \\ \dfrac{dy}{dx} = 2c \ x - 1 \tag{1} \\ \end{gather*}$

Finding the parameter c from the given equation. Then we obtain

$\begin{gather*} x + y = c \ x^2\\ c = \dfrac{x + y}{x^2} \tag{2} \\ \end{gather*}$

Substituting from equation (2) into equation (1), then we get the differential equation of the given family.

$\begin{gather*} \dfrac{dy}{dx} = 2 \ \left(\dfrac{x + y}{x^2}\right) \ x - 1 \\ \dfrac{dy}{dx} = \dfrac{2x + 2y}{x} - 1 \\ \dfrac{dy}{dx} = \dfrac{2x + 2y - x}{x} \\\\ \dfrac{dy}{dx} = \dfrac{x + 2y}{x} \end{gather*}$

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