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Question Number 42446 by ajfour last updated on 25/Aug/18

Commented by ajfour last updated on 26/Aug/18

$F i n d m a x i m u m r i n t e r m s o f$ $a a n d R .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/Aug/18

	$s o l v e x^{2} + {(y - R)}^{2} = R^{2} a n d y = {a x}^{2}$ $x^{2} + y^{2} - 2 y R = 0$ $x^{2} + a^{2} x^{4} - 2 {a x}^{2} R = 0$ $x^{2} (1 + a^{2} x^{2} - 2 a R) = 0$ $a^{2} x^{2} = 2 a R - 1$ $x = \frac{\pm \sqrt{2 a R - 1}}{a}$ $y = {a x}^{2} = a (\frac{2 a R - 1}{a^{2}}) = \frac{2 a R - 1}{a}$ $p o i n t o f i n t e r s e c t i o n a r e$ $(\frac{\sqrt{2 a R - 1}}{a}, \frac{2 a R - 1}{a}) a n d (\frac{- \sqrt{2 a R - 1}}{a}, \frac{2 a R - 1}{a})$ $c o n s i d e r i n g + v e s i d e$ $n o w a r e a b e t w e e n y = {a x}^{2} a n d x^{2} + y^{2} - 2 y R = 0$ $a n d y a x i s s a y = A r e m a i n i n g a r e a b e t w e e n$ $p a r a b o l a a n d c i r c l e = B$ $s o 2 A + 2 B = Π R^{2}$ $B = \frac{Π R^{2}}{2} - A$ $n o w a r e a B = Π r^{2} + r e m a i n i n g a r e a s y m e t r i c a l$ $w . r . t Π r^{2}$ $B = Π r^{2} + 2 S$ $t o m a x i m i z e B$ $c o n t d$
	Answered by MJS last updated on 27/Aug/18

	$tangents parallel but not of interest$ $normals in tangential points must be the same$ $distance between these points = 2 r$ $y = {a x}^{2}$ $T_{p} = (\begin{matrix} p \\ {a p}^{2} \end{matrix})$ $n_{p} : y = k x + d$ $k = - \frac{1}{y^{'}} = - \frac{1}{2 a p} d = \frac{2 a^{2} p^{2} + 1}{2 a}$ $n_{p} : - \frac{1}{2 a p} x + \frac{2 a^{2} p^{2} + 1}{2 a}$ $y = R - \sqrt{R^{2} - x^{2}}$ $T_{c} = (\begin{matrix} q \\ R - \sqrt{R^{2} - q^{2}} \end{matrix})$ $n_{c} : y = k x + d$ $k = - \frac{1}{y^{'}} = - \frac{\sqrt{R^{2} - q^{2}}}{q} d = R$ $n_{c} : y = - \frac{\sqrt{R^{2} - q^{2}}}{q} x + R$ $n_{p} = n_{c}$ $(1) - \frac{1}{2 a p} = - \frac{\sqrt{R^{2} - q^{2}}}{q}$ $(2) \frac{2 a^{2} p^{2} + 1}{2 a} = R$ $\Rightarrow p = - \frac{\sqrt{2 (2 a R - 1)}}{2 a} \land q = - R \sqrt{\frac{2 (2 a R - 1)}{4 a R - 1}}$ $T_{p} = (\begin{matrix} - \frac{\sqrt{2 (2 a R - 1)}}{2 a} \\ \frac{2 a r - 1}{2 a} \end{matrix}) T_{c} = (\begin{matrix} - R \sqrt{\frac{2 (2 a R - 1)}{4 a R - 1}} \\ R (\frac{- 1 + \sqrt{4 a R - 1}}{\sqrt{4 a R - 1}}) \end{matrix})$ $r = \frac{∣ T_{p} T_{c} ∣}{2} = \frac{\sqrt{a R (a R + 1 - \frac{1}{4 a R} - \sqrt{4 a R - 1})}}{2 a}$ $please check, if the idea is ok . . .$
	Commented by ajfour last updated on 27/Aug/18

	$N o s i r, i h a d c o m m i t e d s o m e e r r o r;$ $i h a v e n o t b e e n a b l e t o s o l v e i t y e t . .$
	Commented by MJS last updated on 27/Aug/18

	$ok I^{'} ll check mine . . .$
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