The mirror image of parabola $y^{2} = 4 x$ in tangent to the parabola at (1, 2) is

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${(x + 1)}^{2} = 4 (y - 1)$

${(x - 1)}^{2} = 4 (y - 1)$

${(x - 1)}^{2} = 4 (y + 1)$

${(x + 1)}^{2} = 4 (y + 1)$

answer is A.

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Detailed Solution

$Given the equation of parabola is y^{2} = 4 x Here a = 1 Let any point on the given parabola is (t^{2}, 2 t) . Slope of tangent at (1, 2) = {(\frac{dy}{dx})}_{(1, 2)} = 1 The equation of the tangent at (1, 2) is y = x + 1 \Rightarrow x - y + 1 = 0 The image (h, k) of the point (t^{2}, 2 t) in x - y + 1 = 0 is given by \frac{h - t^{2}}{1} = \frac{k - 2 t}{- 1} = - \frac{2 (t^{2} - 2 t + 1) }{1 + 1} \Rightarrow h - t^{2} = - t^{2} + 2 t - 1 and k - 2 t = t^{2} - 2 t + 1 \Rightarrow h = 2 t - 1 and k = t^{2} + 1 Substitute the value of t in h = 2 t - 1 in k, we get {(h + 1)}^{2} = 4 (k - 1) The required equation of reflection is {(x + 1)}^{2} = 4 (y - 1)$