At which of the following point/s is/are the tangent parallel to the y-axis on the curve $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ ?

(-1, 2)

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(3, 2)

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(1, 4)

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(3, -2)

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Solution

The correct option is B (3, 2)
$Let P (a, b) be the required point . As P (a, b) lies on the curve x^{2} + y^{2} - 2 x - 4 y + 1 = 0 . ∴ a^{2} + b^{2} - 2 a - 4 b + 1 = 0 . . . . . (1] If the tangent to the given curve at P is parallel to y - axis, then {(\frac{dx}{dy})}_{P} = 0 The equation of the curve is x^{2} + y^{2} - 2 x - 4 y + 1 = 0 . Differentiating both sides w . r . t . y, we get 2 x (\frac{dx}{dy}) + 2 y - 2 (\frac{dx}{dy}) - 4 = 0 \Rightarrow 2 \frac{dx}{dy} (x - 1) = 2 (2 - y) \Rightarrow \frac{dx}{dy} = \frac{(2 - y)}{(x - 1)} As {(\frac{dx}{dy})}_{P} = 0 \Rightarrow \frac{(2 - b)}{(a - 1)} = 0 \Rightarrow b = 2 Putting b = 2 in (1] we get a^{2} + 4 - 2 a - 8 + 1 = 0 \Rightarrow a^{2} - 2 a - 3 = 0 \Rightarrow (a + 1) (a - 3) = 0 \Rightarrow a = - 1, 3 Hence, the required points are (- 1, 2) and (3, 2) .$

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