The trajectories orthogonal to $x^{2} + y^{2} = 2 a x$ is

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$y = x^{2}$

$y = C (x^{2} + y^{2})$

$y = C (x^{2} + 2 y^{2})$

$y^{2} = C x$

answer is B.

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Detailed Solution

$2 x + 2 y \frac{d y}{d x} = 2 a \Rightarrow a = x + y \frac{d y}{d x}$

Eliminating $a$ , we have

$x^{2} + y^{2} = 2 (x + y \frac{d y}{d x}) x$

Replacing $\frac{d y}{d x}$ by $- \frac{d x}{d y}$

$\begin{array}{l} x^{2} + y^{2} = 2 (x^{2} - x y \frac{d x}{d y}) \\ \Rightarrow \frac{y^{2} - x^{2}}{- 2 x y} = \frac{d x}{d y} \end{array}$

$\Rightarrow \frac{d y}{d x} = \frac{2 x y}{x^{2} - y^{2}}$ Putting $y = V x$ , we have

$\begin{aligned} V + x \frac{d V}{d x} = \frac{2 V}{1 - V^{2}} \\ \frac{1 - V^{2}}{V (1 + V^{2})} d V = \frac{d x}{x} \\ [\frac{1}{V} - \frac{2 V}{1 + V^{2}}] d V = \frac{d x}{x} \\ \Rightarrow \frac{V}{x (1 + V^{2})} = const \\ \Rightarrow y = C (x^{2} + y^{2}) . \end{aligned}$