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

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$y = x^{2}$

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

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

$y^{2} = x$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n x^{2} + y^{2} = 2 a x - - (1) \\ 2 x + 2 y \frac{d y}{d x} = 2 a \\ x + y \frac{d y}{d x} = a - - (2) \\ P u t (2) i n (1) \\ x^{2} + y^{2} = 2 x (x + y \frac{d y}{d x}) \\ \Rightarrow x^{2} + y^{2} = 2 x^{2} + 2 x y \frac{d y}{d x} \\ \Rightarrow y^{2} - x^{2} = 2 x y \frac{d y}{d x} \\ Re p l a c e \frac{d y}{d x} = - \frac{d x}{d y} \\ \Rightarrow y^{2} - x^{2} = 2 x y (\frac{- d x}{d y}) \\ \Rightarrow \frac{d x}{d y} = \frac{x^{2} - y^{2}}{2 x y} \end{array}$

$\begin{array}{l} P u t x = v y \Rightarrow \frac{d x}{d y} = v + y . \frac{d v}{d y} \\ ∴ v + y . \frac{d v}{d y} = \frac{v^{2} y^{2} - y^{2}}{2 v y^{2}} \\ y \frac{d v}{d y} = \frac{v^{2} - 1}{2 v} - v \\ y . \frac{d v}{d y} = \frac{v^{2} - 1 - 2 v^{2}}{2 v} \\ \int \frac{2 v}{v^{2} + 1} = - \int \frac{1}{y} d y \\ \Rightarrow \log (v^{2} + 1) = - \log y + \log c \\ \Rightarrow y (v^{2} + 1) = c \\ \Rightarrow y (\frac{x^{2}}{y^{2}} + 1) = c \\ \Rightarrow x^{2} + y^{2} = c y \end{array}$