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Question Number 123575 by bramlexs22 last updated on 26/Nov/20

Commented by liberty last updated on 26/Nov/20
$(i) grad normal a circle at point P(a,(√(4−a^2 ))) is m_1 = ((√(4−a^2 ))/a) then eq of normal line ⇒y=((√(4−a^2 ))/a)(x−a)+(√(4−a^2 )) (ii) grad normal a parabola at point Q(b,1+(b−5)^2 ) is m_2 = −(1/(2b−10)) and eq of normal line is ⇒y=−(((x−b))/(2b−10))+1+(b−5)^2 Thus the line is same. we get { ((((√(4−a^2 ))/a) = (1/(10−2b)))),((0 = (b/(2b−10))+1+(b−5)^2 )) :} ⇒ ((3b−10)/(2b−10))+(b−5)^2 = 0 ⇒2b^3 −30b^2 +153b−260=0 (b−4)(2b^2 −22b+65)=0 . this only b=4∈R so we have (1/(10−8)) = ((√(4−a^2 ))/a) ⇒(a^2 /4) = 4−a^2 ; a=± (4/( (√5))) The shortest distance we get from relation d_(min) = OQ−Radius OQ =(√(4^2 +2^2 )) = 2(√5) . Hence d_(min) = 2(√5)−2. ▲$
$(i) g r a d n o r m a l a c i r c l e a t p o i n t$ $P (a, \sqrt{4 - a^{2}}) i s m_{1} = \frac{\sqrt{4 - a^{2}}}{a} t h e n$ $e q o f n o r m a l l i n e \Rightarrow y = \frac{\sqrt{4 - a^{2}}}{a} (x - a) + \sqrt{4 - a^{2}}$ $(i i) g r a d n o r m a l a p a r a b o l a a t p o i n t$ $Q (b, 1 + {(b - 5)}^{2}) i s m_{2} = - \frac{1}{2 b - 10}$ $a n d e q o f n o r m a l l i n e i s \Rightarrow y = - \frac{(x - b)}{2 b - 10} + 1 + {(b - 5)}^{2}$ $T h u s t h e l i n e i s s a m e . w e g e t$ ${\begin{cases} \frac{\sqrt{4 - a^{2}}}{a} = \frac{1}{10 - 2 b} \\ 0 = \frac{b}{2 b - 10} + 1 + {(b - 5)}^{2} \end{cases}$ $\Rightarrow \frac{3 b - 10}{2 b - 10} + {(b - 5)}^{2} = 0$ $\Rightarrow 2 b^{3} - 30 b^{2} + 153 b - 260 = 0$ $(b - 4) (2 b^{2} - 22 b + 65) = 0 . t h i s o n l y$ $b = 4 \in R s o w e h a v e \frac{1}{10 - 8} = \frac{\sqrt{4 - a^{2}}}{a}$ $\Rightarrow \frac{a^{2}}{4} = 4 - a^{2}; a = \pm \frac{4}{\sqrt{5}}$ $T h e s h o r t e s t d i s t a n c e w e g e t f r o m$ $r e l a t i o n d_{m i n} = O Q - R a d i u s$ $O Q = \sqrt{4^{2} + 2^{2}} = 2 \sqrt{5} . H e n c e$ $d_{m i n} = 2 \sqrt{5} - 2 . ▴$
	Answered by MJS_new last updated on 26/Nov/20

	$we need the point of the parabola with the$ $shortest distance to the center of the circle .$ $the center of the circle is (\begin{matrix} 0 \\ 0 \end{matrix}), its radius is 2$ $the points of the parabola are (\begin{matrix} x \\ {(x - 5)}^{2} + 1 \end{matrix})$ $\Rightarrow$ $the square of the distance to (\begin{matrix} 0 \\ 0 \end{matrix}) is$ ${(x)}^{2} + {({(x - 5)}^{2} + 1)}^{2} =$ $= x^{4} - 20 x^{3} + 153 x^{2} - 520 x + 676$ $\frac{d}{d x} [x^{4} - 20 x^{3} + 153 x^{2} - 520 x + 676] = 0$ $4 x^{3} - 60 x^{2} + 306 x - 520 = 0$ $\Rightarrow x = 4$ $\Rightarrow distance is 2 \sqrt{5} to the origin and 2 \sqrt{5} - 2$ $to the line of the circle$ $answer 2 \sqrt{5} - 2$
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