The orthogonal trajectory of the family of parabolas $y^{2} = 4 a x$ is:

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

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

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

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

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Solution

The correct option is C $2 x^{2} + y^{2} = c^{2}$
Given the equation of the family of parabolas is $y^{2} = 4 a x$

Here the parameter is $a$ , which is also an arbitrary constant for finding the ordinary differential equation.

Now differentiating the equation with respectto $x$ on both sides gives,
$\frac{d y}{d x} = \frac{2 a}{y}$

$∴ a = \frac{y}{2} (\frac{d y}{d x})$

substituting in the equation of the family of curves gives,

$y^{2} = 2 x y (\frac{d y}{d x})$ which is differential equation of the family of parabolas.

Now,to find the equation of the orthogonal trajectories we need to replace $(\frac{d y}{d x})$ by $(\frac{- d x}{d y})$ and we need to solve it back
$y^{2} = 2 x y (\frac{- d x}{d y})$

Regrouping the terms and integrating gives,

$\int y d y = \int (- 2 x) d x$

$⟹ \frac{y^{2}}{2} = - x^{2} + c$ where $c$ is the integration constant

regrouping the terms gives,
$2 x^{2} + y^{2} = C^{2}$ where $C$ is a constant.

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The orthogonal trajectory of y^{2} = 4 ax is

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Two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0 a n d x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c^{2} = 0$ are said to be orthogonal. Then $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$

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has?

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The orthogonal trajectory of the family of parabolas y2=4ax is:

The orthogonal trajectory of the family of parabolas $y^{2} = 4 a x$ is: