kerrum75

2021-12-30

Find the orthogonal trajectory of the family of curves $y=c{x}^{2}$.

Piosellisf

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

Replacing

$\frac{{y}^{2}}{2}=-\frac{{x}^{2}}{2×2}+\frac{{A}^{2}}{4}$
$\frac{2{y}^{2}}{4}=-\frac{{x}^{2}}{4}+\frac{{A}^{2}}{4}$
$\frac{2{y}^{2}}{\left\{4\right\}}=\left\{\frac{1}{4}\right\}\left[-{x}^{2}+{A}^{2}\right]$
${y}^{2}=\frac{1}{2}\left[-{x}^{2}+{A}^{2}\right]$
$2{y}^{2}=-{x}^{2}+{A}^{2}$

Answer ${x}^{2}+2{y}^{2}={A}^{2}$ where A is constant.

yotaniwc

$y=C{x}^{2}$
$⇒\frac{y}{{x}^{2}}=C$,
Differentiating $\frac{{x}^{2}{y}^{\prime }-2xy}{{x}^{4}}=0$
$⇒{y}^{\prime }=\frac{2y}{x}$
Replacing , we have ${y}^{\prime }=\frac{-x}{2y}$
$⇒2ydy+xdx=0$
$⇒{y}^{2}+\frac{1}{2}{x}^{2}=\text{constant}$
which represent a family of ellipses.

Vasquez

$y=c{x}^{2}$ (1)
differentiating (1) x, we get
${y}^{\prime }=2cx$ (2)
From (1) $c=\frac{y}{{x}^{2}}$, using in (2), we get
${y}^{\prime }=\frac{2yx}{{x}^{2}}⇒{y}^{\prime }=\frac{2y}{x}$ (3)
Since, slopes of arthogonal trajectories are negative reciprocals of each other, then orthogonal trajectory is given by solving: ${y}^{\prime }=-\frac{x}{2y}\to$ Slope of orthogonal (4) trajectories
Solving (4), $\frac{dy}{dx}=\frac{-x}{2y}$
$⇒\int 2ydy=\int -xdx$
$⇒{y}^{2}=-\frac{{x}^{2}}{2}+{K}_{1}$
or $2{y}^{2}+{x}^{2}=K;K=2{K}_{1}$
So (5) represent arthogonal trajectories of (1)

Do you have a similar question?