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

\sqrt{21}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{26} - \sqrt{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{20} - \sqrt{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{21} - \sqrt{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{20} - \sqrt{5}$
Let normal to the parabola $y^{2} = 4 x$ at point $P (t^{2}, 2 t)$ be $y + t x = 2 t + t^{3}$
for shortest distance it must pass through centre of the given circle i.e $C (6, 0)$
$\Rightarrow 6 t = 2 t + t^{3}$ .
$\Rightarrow t = 2$
Thus point P is $(4, 4)$
Hence shortest distance $= C P - r = \sqrt{4 + 16} - \sqrt{5} = \sqrt{20} - \sqrt{5}$

Suggest Corrections

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

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

Join BYJU'S Learning Program