Minimum distance between the curves $y^{2} = 4 x$ and $x^{2} + y^{2} - 12 x + 31 = 0$ is

\sqrt{5}

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\sqrt{21}

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\sqrt{28} - \sqrt{5}

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\sqrt{21} - \sqrt{5}

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Solution

The correct option is A $\sqrt{5}$
Centre and radius of the given circle is $P (6, 0)$ and $\sqrt{5}$ , respectively.

Now the minimum distance between two curves always occurs along a line that normal to both the curves.

Equation of normal to $y^{2} = 4 x$ at $(t^{2}, 2 t)$ is

$y = - t x + 2 t + t^{3}$ .

If it is normal to circle also, then it must pass through $(6, 0)$ .

$∴ 0 = t^{3} - 4 t \Rightarrow t = 0$ or $t = \pm 2$ .

$\Rightarrow A (4, 4)$ and $C (4, - 4)$

$\Rightarrow P A = P C = \sqrt{20} = 2 \sqrt{5}$

Required minimum distance $= 2 \sqrt{5} - \sqrt{5} = \sqrt{5}$ .

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Similar questions

Minimum distance between the curve $y^{2} = 4 x$ and $x^{2} + y^{2} - 12 x + 31 = 0$ is

x^{2} + y^{2} + 4 x + 16 y + 66 = 0

y^{2} = 8 x

y^{2} = 4 x

x^{2} + y^{2} = 5

S_{1} \equiv x^{2} + y^{2} - 4 y - 5 = 0

S_{2} \equiv x^{2} + y^{2} + 12 x + 4 y + 31 = 0

S_{2} \equiv x^{2} + y^{2} + 6 x + 12 y + 36 = 0

y^{2} = 4 x, x^{2} + y^{2} = 5

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