Find the minimum distance between C_f , C_g ; C_f : y^2 = 4ax , C


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Question Number 183324 by Acem last updated on 25/Dec/22

$F i n d t h e m i n i m u m d i s t a n c e b e t w e e n C_{f}, C_{g}$ $; C_{f} : y^{2} = 4 a x, C_{g} : x^{2} + y^{2} - 24 a y + 128 a^{2} = 0$
Commented by Acem last updated on 25/Dec/22

$N o m a t t e r i f \forall a \in R^{* +} o r R^{* -}$
	Answered by mr W last updated on 25/Dec/22

	Commented by mr W last updated on 25/Dec/22

	Commented by mr W last updated on 25/Dec/22

	$a \neq 0, a s s u m e a > 0$ $y^{2} = 4 a x$ $x^{2} + {(y - 12 a)}^{2} = {(4 a)}^{2}$ $A (0, 12 a)$ $r = 4 a$ $s a y P (\frac{p^{2}}{4 a}, p)$ $\tan θ = \frac{d y}{d x} = \frac{2 a}{y} = \frac{2 a}{p} = \frac{\frac{p^{2}}{4 a}}{12 a - p}$ $\Rightarrow {(\frac{p}{a})}^{3} + 8 (\frac{p}{a}) - 96 = 0$ $\Rightarrow \frac{p}{a} = 4$ $R = A P = \sqrt{{(\frac{p^{2}}{4 a})}^{2} + {(12 a - p)}^{2}} = 4 \sqrt{5} ∣ a ∣$ $d_{m i n} = R - r = 4 (\sqrt{5} - 1) ∣ a ∣$
	Commented by Acem last updated on 25/Dec/22

	$T h x S i r!$
	Answered by Acem last updated on 25/Dec/22

	$F o r a > 0$ $C_{c i r c l e} (0, 12 a), R = 4 a$ $y = 2 \sqrt{a} \sqrt{x}^{T h e u p p e r p a r t o f t h e p a r a b o l a}$ $A s s u m e t h a t x = {a k}^{2} t h e n y = 2 a k, P ({a k}^{2}, 2 a k) \in C_{f}$ $T h e n o r m a l △ p a s s e s t h r o u g h P a n d t h e c e n t e r o f t h e c i r c l e C$ $△ : y + x k = 2 a k + {a k}^{3} = 12 a_{x_{c} = 0} \Leftrightarrow k^{3} + 2 k - 12 = 0$ $\Rightarrow (k - 2) \underset{N o s o l u t i o n i n R}{\underset{⏟}{(k^{2} + 2 k + 6)}} = 0 \Rightarrow k = 2$ $\Rightarrow P (4 a, 4 a), T \in C_{g}, △$ $∣ T P ∣_{m i n . D i s .} = \sqrt{{(- 4 a)}^{2} + {(12 a - 4 a)}^{2}} - 4 a = 4 a (\sqrt{5} - 1) u n i .$ $; ∣ T P ∣= ∣ C P ∣ - R$
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