"Find the orthogonal trajectory of the family of..." solved and explained

kerrum75

Answered question

2021-12-30

Find the orthogonal trajectory of the family of curves

y = c x^{2}

Answer & Explanation

Piosellisf

Beginner2021-12-31Added 40 answers

$y = c x^{2}$
$\frac{d y}{d x} = 2 c x$
$c = \frac{1}{2 x} (\frac{d y}{d x})$
$y = \frac{1}{2 x} (\frac{d y}{d x}) \times x^{2}$
$y = \frac{x}{2} (\frac{d y}{d x})$
Replacing $\frac{d y}{d x} w i t h \frac{- 1}{(\frac{d y}{d x})}$
$y = \frac{x}{2} (\frac{- 1}{\frac{d y}{d x}})$
$- (\frac{d y}{d x}) \times y = \frac{x}{2}$
$y d y = - \frac{x}{2} d x$
$\int y d y = - \int \frac{x}{2} d x$
$\frac{y^{2}}{2} = - \frac{x^{2}}{2 \times 2} + \frac{A^{2}}{4}$
$\frac{2 y^{2}}{4} = - \frac{x^{2}}{4} + \frac{A^{2}}{4}$
$\frac{2 y^{2}}{{4}} = {\frac{1}{4}} [- x^{2} + A^{2}]$
$y^{2} = \frac{1}{2} [- x^{2} + A^{2}]$
$2 y^{2} = - x^{2} + A^{2}$

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

yotaniwc

Beginner2022-01-01Added 34 answers

y = C x^{2}

\Rightarrow \frac{y}{x^{2}} = C

,
Differentiating

\frac{x^{2} y^{'} - 2 x y}{x^{4}} = 0

\Rightarrow y^{'} = \frac{2 y}{x}

Replacing

\frac{d y}{d x} b y - \frac{d x}{d y}

, we have

y^{'} = \frac{- x}{2 y}

\Rightarrow 2 y d y + x d x = 0

\Rightarrow y^{2} + \frac{1}{2} x^{2} = constant

which represent a family of ellipses.

Vasquez

Expert2022-01-09Added 669 answers

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

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