Minimum-distance-between-curves-y-2-4x-and-x-2-y-2-12x-31-0-is-

Question Number 57578 by rahul 19 last updated on 07/Apr/19

M i n i m u m d i s t a n c e b e t w e e n c u r v e s

y^{2} = 4 x a n d x^{2} + y^{2} - 12 x + 31 = 0 i s ?

Answered by mr W last updated on 07/Apr/19

x^{2} + y^{2} - 12 x + 31 = 0

{(x - 6)}^{2} + y^{2} = {(\sqrt{5})}^{2}

\Rightarrow c i r c l e w i t h r a d i u s \sqrt{5} a n d c e n t e r (6, 0) .

e q n . o f c i r c l e w i t h r a d i u s r a n d c e n t e r (6, 0) :

{(x - 6)}^{2} + y^{2} = r^{2}

s u c h t h a t t h e c i r c l e t o u c h e s t h e c u r v e

y^{2} = 4 x

\Rightarrow {(x - 6)}^{2} + 4 x = r^{2}

\Rightarrow x^{2} - 8 x + (36 - r^{2}) = 0

\Rightarrow Δ = 64 - 4 (36 - r^{2}) = 0

\Rightarrow r = 2 \sqrt{5}

M i n i m u m d i s t a n c e b e t w e e n c u r v e s

y^{2} = 4 x a n d x^{2} + y^{2} - 12 x + 31 = 0 i s

2 \sqrt{5} - \sqrt{5} = \sqrt{5}

Commented by mr W last updated on 07/Apr/19

Commented by rahul 19 last updated on 08/Apr/19

thank you sir.

Answered by MJS last updated on 07/Apr/19

the center of the circle is (\begin{matrix} 6 \\ 0 \end{matrix}); r = \sqrt{5}

\Rightarrow the point of the parabola with minimum

distance from this center is the same as the

point with the minimum distance of the

circle .

∣ [\begin{matrix} x \\ 2 \sqrt{x} \end{matrix}] - [\begin{matrix} 6 \\ 0 \end{matrix}] ∣^{2} = x^{2} - 8 x + 36

\frac{d}{d x} [x^{2} - 8 x + 36] = 2 x - 8 = 0 \Rightarrow x = 4

distance = \sqrt{x^{2} - 8 x + 36} - \sqrt{5} = \sqrt{20} - \sqrt{5} = \sqrt{5}

Commented by rahul 19 last updated on 08/Apr/19

thank you sir.