Find-minimum-distance-between-y-2-8x-and-x-2-y-6-2-1-

Question Number 37137 by rahul 19 last updated on 09/Jun/18

Find minimum distance between

y^{2} = 8 x a n d x^{2} + {(y + 6)}^{2} = 1 .

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

f r o m g r a p h i t i s s e e n t h a t (2, - 4) l i e s o n y^{2} = 8 x

(2, - 4) i s n e a r e s t p o i n t t o t h e c i r c l e

d i s t a n c e b e t w e e n c e n t r e o f c i r c l e (0, - 6) a n d

(2, - 4 i s

\sqrt{{(2 - 0)}^{2} + {(- 4 + 6)}^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}

s o m i n i m u m d i s t a n c e = 2 \sqrt{2} - 1

p l s c h e c k

Commented by rahul 19 last updated on 09/Jun/18

Sir, how do you conclude (2, - 4) is the

nearest point to circle .

Also sol . using parametric coordinates

is welcome!

Commented by rahul 19 last updated on 09/Jun/18

I have made this random problem myself

so i {don}^{'} t have any answer .

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18

(2 t^{2}, 4 t) l i e z o n y^{2} = 8 x

d i s t a n c e b e t w e e n (2 t^{2}, 4 t) a n d t h e c e n t r e (0, - 6)

t o b e m i n i m u m

s = \sqrt{{(2 t^{2} - 0)}^{2} + {(4 t + 6)}^{2}}

s^{2} = 4 t^{4} + 16^{2} + 48 t + 36

s^{2} = 4 (t^{4} + 4 t^{2} + 12 t + 9)

2 s \frac{d s}{d t} = 4 (4 t^{3} + 8 t + 12)

f o r m i n o r m a x

\frac{d s}{d t} = 0 = 4 t^{3} + 8 t + 12

t^{3} + 2 t + 3 = 0

= t^{3} + t^{2} - t^{2} - t + 3 t + 3

= t^{2} (t + 1) - t (t + 1) + 3 (t + 1)

= (t + 1) (t^{2} - t + 3)

s \frac{d s}{d t} = 2 (4 t^{3} + 8 t + 12)

s \frac{d s}{d t} = 8 (t^{3} + 2 t + 3)

s \frac{d^{2} s}{{d t}^{2}} + {(^{} \frac{d s}{d t})}^{2} = 8 (3 t^{2} + 2)

p u t t = - 1 i n 8 (3 t^{2} + 2) i s 40 > 0

+ v e s o

m i n d i s t a n c e = \sqrt{2^{2} + {(- 4 + 6)}^{2}} f r o m c e n t r e

= \sqrt{8} = 2 \sqrt{2}

r e q u i r e d d i s t a n c e = 2 \sqrt{2} - 1

o

Answered by MJS last updated on 10/Sep/18

the minimum distance between the circle and

the parabola is the mininum distance between

the center of the circle and the parabola

minus the radius of the circle

P \in p a r (negative branch) : P = (\begin{matrix} x \\ - 2 \sqrt{2 x} \end{matrix})

center of circle : C = (\begin{matrix} 0 \\ - 6 \end{matrix})

∣ P C ∣^{2} = x^{2} + {(- 6 + 2 \sqrt{2 x})}^{2} = x^{2} + 8 x - 24 \sqrt{2 x} + 36

\frac{d}{d x} [x^{2} + 8 x - 24 \sqrt{2 x} + 36] = 0

2 x - \frac{12 \sqrt{2 x}}{x} + 8 = 0 \Rightarrow

\Rightarrow x (x - 2) (x^{2} + 10 x + 36) = 0

x_{1} = 0 is a local maximum

x_{2} = 2 is a local minimum

P = (\begin{matrix} 2 \\ - 4 \end{matrix})

∣ P C ∣= \sqrt{x^{2} + 8 x - 24 \sqrt{2 x} + 36} = 2 \sqrt{2}

minimum distance =∣ P C ∣ - r = 2 \sqrt{2} - 1

Commented by ajfour last updated on 09/Jun/18

G r e a t! S i r .

(b u t w h y d i d y o u d e l e t e y o u r p o s t

o f s e v e r a l i n t e g r a t i o n q u e s t i o n s ?)

Commented by MJS last updated on 09/Jun/18

I made some mistakes, will post the corrected

version soon

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