## Related questions with answers

Find the orthogonal trajectories of each given family of curves. In each case sketch several members of the family and several of the orthogonal trajectories on the same set of axes.

$y^{2}=c x$

Solution

VerifiedDifferentiate the given equation

$y^2=cx$

with respect to $x$ , we get

$2y\frac{dy}{dx}=c \hspace*{15mm} (1)$

Finding the parameter $c$ from the given equation. Then we get

$c=\frac{y^2}{x}$

Substitute $c$ into the equation $(1)$ , we get

$\begin{align*} 2y\frac{dy}{dx}&=\frac{y^2}{x}\\ \frac{dy}{dx}&=\frac{y}{2x} \end{align*}$

Now, find the differential equation of the orthogonal trajectories by replacing $\frac{y}{2x}$ by its negative reciprocal $-2\frac{x}{y}.$ Then we get

$\begin{align*} \frac{dy}{dx}&=-2\frac{x}{y}\\ y\, dy &=-2x\, dx \end{align*}$

By integration, then we see that

$\begin{align*} \int y\, dy&=\int -2x\, dx\\ \frac{y^2}{2} &=-2\int x\, dx\\ \frac{y^2}{2}&=-x^2+c_1\\ c&=\boxed{y^2+2x^2} \end{align*}$

Therefore

$\boxed{\color{#4257b2} y^2+2x^2=c}$

is the family of orthogonal trajectories of the given family.

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