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

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 D $1$
If normals are drawn from a point $(h, k)$ to a parabola $y^{2} = 4 a x$
1) There are $3$ normals if $h > 2 a$
2) There is only one normal if $h < 2 a$
So,
In given equation,
$y^{2} - 2 y - 4 x - 3 = 0$
$({y - 1)}^{2} = 4 (x + 1)$
here $a = 2$ $\Rightarrow 2 a = 4$
point $(h, k)$ $= (- 2, 1)$
$∴ - 2 < 4$
Hence, there is only $1$ normal.

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(2, 1)

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

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y^{2} = 4 a x, (a > 0)

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