The number of distinct normals that can be drawn from $(- 2, 1)$ to the parabola $y^{2} - 4 x - 2 y - 3 = 0$ is/are :

1

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2

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3

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0

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Solution

The correct option is A $1$
Since, $y^{2} - 4 x - 2 y - 3 = 0$
i.e., $(y^{2} - 2 y + 1) - 4 x - 4 = 0$
i.e., $(y - 1)^{2} = 4 + 4 x$
i.e., $(y - 1)^{2} = 4 (x + 1)$
i.e., The vertex of parabola will lie at $(- 1, 1)$
And the point $(- 2, 1)$ lie parallel to the vertex point $(- 1, 1)$
Hence, only one normal can be drawn from $(- 2, 1)$ to the parabola $y^{2} - 4 x - 2 y - 3 = 0$

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(2, 1)

y^{2} - 4 x - 2 y - 3 = 0

(- 2, 1)

y^{2} - 4 x - 2 y - 3 = 0

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