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Question Number 72721 by mr W last updated on 01/Nov/19

Commented by mr W last updated on 01/Nov/19

$f i n d t h e c o r r e l a t i o n o f A a n d B s u c h$ $t h a t t h e c i r c l e i s i n s c r i b e d b e t w e e n t h e$ $p a r a b o l a a n d t h e x - a x i s a s s h o w n .$
Commented by kaivan.ahmadi last updated on 01/Nov/19

$(d, 2 R) \in y = A x + {B x}^{2} \Rightarrow$ $2 R = A d + {B d}^{2} \Rightarrow A = \frac{2 R - {B d}^{2}}{d}$
Commented by mr W last updated on 01/Nov/19

$n o t c o r r e c t s i r!$ $t h e p a r a b o l a s h o u l d j u s t t a n g e n t t h e$ $c i r c l e .$
Commented by mr W last updated on 01/Nov/19

Commented by kaivan.ahmadi last updated on 01/Nov/19

$i n t h i s c a s e f i r s t w e h a v a {(x - d)}^{2} + {(y - R)}^{2} = R^{2}$ $a n d y = A x + {B x}^{2} .$ $w e m u s t f i n d t h e r o o t s o f e q u a t i o n$ ${(x - d)}^{2} + {(A x + {B x}^{2} - R)}^{2} = R^{2}$
Commented by ajfour last updated on 01/Nov/19

Commented by ajfour last updated on 02/Nov/19

$y = {a x}^{2}$ ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $\Rightarrow {(x - h)}^{2} + {({a x}^{2} - k)}^{2} = r^{2}$ $w i t h 2 a x = - \frac{x - h}{{a x}^{2} - k}$ $\Rightarrow 2 a^{2} x^{3} + (1 - 2 a k) x - h = 0$ $x^{3} + (\frac{1 - 2 a k}{2 a^{2}}) x - \frac{h}{2 a^{2}} = 0$ $D < 0$ $t h e i n e v i t a b l e r e a l r o o t f r o m$ ${C a r d a n o}^{'} s s o l u t i o n b e x_{0} .$ $s a y x_{0} = f (a, k)$ $\Rightarrow {a x}_{0}^{2} = k - \sqrt{r^{2} - {(x_{0} - h)}^{2}}$ $w i t h k + r = a {(h + d)}^{2}$ $_________________________$ $Y = A X + {B X}^{2}$ $X_{0} = - \frac{A}{2 B} = h + d$ $Y_{m a x} = - \frac{A^{2}}{4 B} = a {(h + d)}^{2} = k + r$ $h e n c e A a n d B c o r e l a t e d,$ $a f t e r p r o p e r s u b s t i t u t i o n s .$
Commented by mr W last updated on 02/Nov/19

$t h a n k s s i r!$ $w e o n l y e n s u r e t h a t p a r a b o l a t a n g e n t s$ $t h e c i r c l e, b u t b o t h c a n i n t e r s e c t a t$ $a n o t h e r p o i n t . w e s h o u l d f i n d t h e$ $c o n d i t i o n t h a t b o t h i n t e r s e c t a t o n e$ $s i n g l e p o i n t o r o n e d o u b l e p o i n t .$
Commented by MJS last updated on 02/Nov/19

Commented by MJS last updated on 02/Nov/19
I once posted this, it might help here. But I'm afraid we cannot generally solve.
Commented by ajfour last updated on 02/Nov/19
thanks for sharing it again, mjs sir, but herein by simply imposing D<0 would suffice for the matter is that of a cubic and not quartic, i guess.
Commented by mr W last updated on 02/Nov/19

$t h a n k y o u f o r t h i s m a t e r i a l s i r!$
	Answered by mr W last updated on 02/Nov/19

	Commented by mr W last updated on 02/Nov/19

	$w e m a y h a v e t w o c a s e s :$ $c a s e 1 : p a r a b o l a t o u c h e s t h e c i r c l e a t$ $t w o p o i n t s$ $c a s e 2 : p a r a b o l a t o u c h e s t h e c i r c l e o n l y$ $a t o n e p o i n t$ $c a s e 1 :$ $y = A x + {B x}^{2} = B [x^{2} + \frac{A}{B} x + {(\frac{A}{2 B})}^{2}] - \frac{A^{2}}{4 B}$ $y = B {(x + \frac{A}{2 B})}^{2} - \frac{A^{2}}{4 B}$ $\Rightarrow \frac{A}{2 B} = - d$ $y = B {(x - d)}^{2} - {B d}^{2} = B (x^{2} - 2 d x)$ $y^{'} = 2 B (x - d)$ $p o i n t B (d - R \sin φ, R (1 + \cos φ))$ $\tan φ = 2 B (d - R \sin φ - d)$ $\Rightarrow B = - \frac{1}{2 R \cos φ}$ $R (1 + \cos φ) = B (d - R \sin φ - 2 d) (d - R \sin φ)$ $R (1 + \cos φ) = \frac{d^{2} - R^{2} \sin^{2} φ}{2 R \cos φ}$ $1 + \cos φ = \frac{δ^{2} - \sin^{2} φ}{2 \cos φ} w i t h δ = \frac{d}{R}$ ${(\cos φ + 1)}^{2} = δ^{2}$ $\Rightarrow \cos φ = δ - 1$ $\Rightarrow B = - \frac{1}{2 R (δ - 1)}$ $i . e . w e h a v e c a s e 1 i f 1 < δ ⩽ 2 .$ $i n t h i s c a s e t h e e q n . o f p a r a b o l a i s$ $y = - \frac{x (x - 2 d)}{2 R (δ - 1)}$ $o r A = - \frac{δ}{δ - 1}, B = - \frac{1}{2 R (δ - 1)}$ $c a s e 2 :$ $y = A x + {B x}^{2}$ $y^{'} = A + 2 B x$ $p o i n t B (d - R \sin φ, R (1 + \cos φ))$ $\tan φ = A + 2 B (d - R \sin φ) . . . (i)$ $R (1 + \cos φ) = (d - R \sin φ) [A + B (d - R \sin φ)]$ $\frac{R (1 + \cos φ)}{(d - R \sin φ)} = A + B (d - R \sin φ) . . . (i i)$ $(i) - (i i) :$ $\Rightarrow B^{'} = B R = \frac{(δ - \sin φ) \tan φ - (1 + \cos φ)}{{(δ - \sin φ)}^{2}}$ $2 (i i) - (i) :$ $\Rightarrow A = \frac{2 (1 + \cos φ) - (δ - \sin φ) \tan φ}{δ - \sin φ}$
	Commented by ajfour last updated on 02/Nov/19

	$t h e s e c o n d c a s e d o e s g e t$ $p a r a m e t r i c, S i r . I t r i e d b u t$ $d i n t p o s t .$
	Commented by mr W last updated on 02/Nov/19

	$b u t w i t h m y m e t h o d t h i n g s c a n o c c u r$ $l i k e t h i s :$
	Commented by mr W last updated on 02/Nov/19

	Commented by ajfour last updated on 02/Nov/19

	$K e e n o b s e r v a t i o n S i r, i j u s t$ $p o s t e d, p l e a s e v i e w . .$
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