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Question Number 213281 by ajfour last updated on 02/Nov/24

Commented by ajfour last updated on 02/Nov/24

$F i n d r i n t e r m s o f a, b .$
	Answered by mr W last updated on 02/Nov/24
	$say touching point from circle and parabola at (±p, p^2 ). tan θ=2p r sin θ=p ⇒r=((√(1+4p^2 ))/( 2)) y_c =p^2 +r cos θ=p^2 +(1/2) eqn. of tangent line: y=−((b^2 (x−a))/(a+b))=((ab^2 )/(a+b))−((b^2 x)/(a+b)) x^2 +(((ab^2 )/(a+b))−((b^2 x)/(a+b))−p^2 −(1/2))^2 =((1+4p^2 )/4) x^2 +(((ab^2 )/(a+b))−p^2 −(1/2)−((b^2 x)/(a+b)))^2 =((1+4p^2 )/4) [1+(b^4 /((a+b)^2 ))]x^2 −((2b^2 )/((a+b)))(((ab^2 )/(a+b))−p^2 −(1/2))x+(((ab^2 )/(a+b))−p^2 −(1/2))^2 −((1+4p^2 )/4)=0 Δ=(b^4 /((a+b)^2 ))(((ab^2 )/(a+b))−p^2 −(1/2))^2 −[1+(b^4 /((a+b)^2 ))][(((ab^2 )/(a+b))−p^2 −(1/2))^2 −((1+4p^2 )/4)]=0 ((1/4)+p^2 )[1+(b^4 /((a+b)^2 ))]−(((ab^2 )/(a+b))−(1/2)−p^2 )^2 =0 p^4 −(b^2 /(a+b))(2a+(b^2 /(a+b)))p^2 +(a^2 +(b^2 /4))((a^2 b^4 )/((a+b)^2 ))−((ab^2 )/(a+b))=0 ⇒p^2 =(1/2){(b^2 /(a+b))(2a+(b^2 /(a+b)))−(√([(2ab+(b^3 /(a+b)))^2 −4a^4 −a^2 b^2 ](b^4 /((a+b)^2 ))+((4ab^2 )/(a+b))))}$
	$s a y t o u c h i n g p o i n t f r o m c i r c l e a n d$ $p a r a b o l a a t (\pm p, p^{2}) .$ $\tan θ = 2 p$ $r \sin θ = p$ $\Rightarrow r = \frac{\sqrt{1 + 4 p^{2}}}{2}$ $y_{c} = p^{2} + r \cos θ = p^{2} + \frac{1}{2}$ $e q n . o f t a n g e n t l i n e :$ $y = - \frac{b^{2} (x - a)}{a + b} = \frac{{a b}^{2}}{a + b} - \frac{b^{2} x}{a + b}$ $x^{2} + {(\frac{{a b}^{2}}{a + b} - \frac{b^{2} x}{a + b} - p^{2} - \frac{1}{2})}^{2} = \frac{1 + 4 p^{2}}{4}$ $x^{2} + {(\frac{{a b}^{2}}{a + b} - p^{2} - \frac{1}{2} - \frac{b^{2} x}{a + b})}^{2} = \frac{1 + 4 p^{2}}{4}$ $[1 + \frac{b^{4}}{{(a + b)}^{2}}] x^{2} - \frac{2 b^{2}}{(a + b)} (\frac{{a b}^{2}}{a + b} - p^{2} - \frac{1}{2}) x + {(\frac{{a b}^{2}}{a + b} - p^{2} - \frac{1}{2})}^{2} - \frac{1 + 4 p^{2}}{4} = 0$ $Δ = \frac{b^{4}}{{(a + b)}^{2}} {(\frac{{a b}^{2}}{a + b} - p^{2} - \frac{1}{2})}^{2} - [1 + \frac{b^{4}}{{(a + b)}^{2}}] [{(\frac{{a b}^{2}}{a + b} - p^{2} - \frac{1}{2})}^{2} - \frac{1 + 4 p^{2}}{4}] = 0$ $(\frac{1}{4} + p^{2}) [1 + \frac{b^{4}}{{(a + b)}^{2}}] - {(\frac{{a b}^{2}}{a + b} - \frac{1}{2} - p^{2})}^{2} = 0$ $p^{4} - \frac{b^{2}}{a + b} (2 a + \frac{b^{2}}{a + b}) p^{2} + (a^{2} + \frac{b^{2}}{4}) \frac{a^{2} b^{4}}{{(a + b)}^{2}} - \frac{{a b}^{2}}{a + b} = 0$ $\Rightarrow p^{2} = \frac{1}{2} {\frac{b^{2}}{a + b} (2 a + \frac{b^{2}}{a + b}) - \sqrt{[{(2 a b + \frac{b^{3}}{a + b})}^{2} - 4 a^{4} - a^{2} b^{2}] \frac{b^{4}}{{(a + b)}^{2}} + \frac{4 {a b}^{2}}{a + b}}}$
	Commented by mr W last updated on 02/Nov/24

	Answered by ajfour last updated on 02/Nov/24
	$let center of circle be origin. y−(b^2 −y_c )=−((b^2 (x+b))/((a+b))) tan α=(b^2 /(a+b)) (r/(cos α))=b^2 −y_c −(b^3 /(a+b)) r=(((a+b))/( (√(b^4 +(a+b)^2 )))){b^2 −y_c −(b^3 /(a+b))} Eq. of parabola y=x^2 −y_c circle eq: x^2 +y^2 =r^2 (x^2 −y_c )^2 +x^2 =r^2 x^4 −(1−2y_c )x^2 +y_c ^2 −r^2 =0 △=0 ⇒ (1−2y_c )^2 =4(y_c ^2 −r^2 ) ⇒ 1−4y_c +4r^2 =0 since r=(((a+b))/( (√(b^4 +(a+b)^2 )))){b^2 −y_c −(b^3 /(a+b))} r=k(((ab^2 )/(a+b))−r^2 −(1/4)) r^2 +(r/k)−((ab^2 )/(a+b))+(1/4)=0 r=−(1/(2k))+(√((1/(4k^2 ))−(1/4)+((ab^2 )/(a+b)))) where k=(((a+b))/( (√(b^4 +(a+b)^2 ))))$
	$l e t c e n t e r o f c i r c l e b e o r i g i n .$ $y - (b^{2} - y_{c}) = - \frac{b^{2} (x + b)}{(a + b)}$ $\tan α = \frac{b^{2}}{a + b}$ $\frac{r}{\cos α} = b^{2} - y_{c} - \frac{b^{3}}{a + b}$ $r = \frac{(a + b)}{\sqrt{b^{4} + {(a + b)}^{2}}} {b^{2} - y_{c} - \frac{b^{3}}{a + b}}$ $E q . o f p a r a b o l a$ $y = x^{2} - y_{c}$ $c i r c l e e q : x^{2} + y^{2} = r^{2}$ ${(x^{2} - y_{c})}^{2} + x^{2} = r^{2}$ $x^{4} - (1 - 2 y_{c}) x^{2} + y_{c}^{2} - r^{2} = 0$ $△ = 0 \Rightarrow {(1 - 2 y_{c})}^{2} = 4 (y_{c}^{2} - r^{2})$ $\Rightarrow 1 - 4 y_{c} + 4 r^{2} = 0$ $s i n c e$ $r = \frac{(a + b)}{\sqrt{b^{4} + {(a + b)}^{2}}} {b^{2} - y_{c} - \frac{b^{3}}{a + b}}$ $r = k (\frac{{a b}^{2}}{a + b} - r^{2} - \frac{1}{4})$ $r^{2} + \frac{r}{k} - \frac{{a b}^{2}}{a + b} + \frac{1}{4} = 0$ $r = - \frac{1}{2 k} + \sqrt{\frac{1}{4 k^{2}} - \frac{1}{4} + \frac{{a b}^{2}}{a + b}}$ $w h e r e k = \frac{(a + b)}{\sqrt{b^{4} + {(a + b)}^{2}}}$
	Commented by ajfour last updated on 02/Nov/24

	$F o r a = 4, b = 2$ $\frac{1}{2 k} = \frac{\sqrt{16 + 36}}{12} = \frac{\sqrt{13}}{6}$ $r = - \frac{\sqrt{13}}{6} + \sqrt{\frac{13 - 9 + 96}{36}}$ $r = \frac{10 - \sqrt{13}}{6} \approx 1.06574$ $y_{c} = \frac{1}{4} + \frac{113 - 20 \sqrt{13}}{36} = \frac{61 - 10 \sqrt{13}}{18}$ $y_{c} \approx 1.3858$
	Commented by ajfour last updated on 02/Nov/24

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