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Question Number 162309 by Mathematification last updated on 28/Dec/21

Commented by mr W last updated on 28/Dec/21

$i t s e e m s t h a t y o u a n d m s . t a w a a r e$ $v i s i t i n g t h e s a m e c o l l e g e, s i n c e y o u r$ $q u e s t i o n s s e e m t o b e f r o m t h e s a m e$ $s o u r c e (o r t e a c h e r) .$
Commented by Mathematification last updated on 28/Dec/21

$T h e q u e s t i o n i s r e a l l y f r o m m y f r i e n d, I s h a r e d$ $i t i n a w h a t s a p p g r o u p, t h a t w a s w h e n m . s t a w a$ $(w h o m i d o n t k n o w) f o u n d t h e q u e s t i o n .$ $I a l s o s a w t h e p r e v i o u s q u e s t i o n s s h e p o s t e d h e r e .$
Commented by mr W last updated on 28/Dec/21

$i s e e . t h a n k s f o r c l a r i f i n g!$
	Answered by mr W last updated on 28/Dec/21

	Commented by Mathematification last updated on 29/Dec/21

	$P l e a s e h o w d i d y o u k n o w t h a t t h e c o o r d i n a t e$ $c o n n e c t i n g t h e p a r a b o l a a n d t h e c i r c u m f e r e n c e$ $i s (p, p^{2}) ? ?$
	Commented by mr W last updated on 28/Dec/21

	$P (p, p^{2})$ $\tan θ = \frac{d y}{d x} = 2 x = 2 p$ $x_{A} = p + R \sin θ$ $y_{A} = p^{2} - R \cos θ$ ${(x_{A} - 7)}^{2} + {(y_{A} - 3)}^{2} = {(3 + R)}^{2} . . . (i)$ ${(x_{A} - 11)}^{2} + {(y_{A} - 3)}^{2} = {(1 + R)}^{2} . . . (i i)$ $(i) - (i i) :$ $4 (2 x_{A} - 18) = 2 (2 R + 4)$ $\Rightarrow x_{A} = \frac{R}{2} + 10$ $\frac{R}{2} + 10 = p + R \sin θ = p + \frac{2 p R}{\sqrt{1 + 4 p^{2}}}$ $\Rightarrow R = \frac{2 (10 - p)}{\frac{4 p}{\sqrt{1 + 4 p^{2}}} - 1}$ $\Rightarrow {(\frac{R}{2} + 3)}^{2} + {(p^{2} - \frac{R}{\sqrt{1 + 4 p^{2}}} - 3)}^{2} = {(3 + R)}^{2}$ $\Rightarrow p \approx 4.0641$ $\Rightarrow R \approx 12.0522$
	Commented by mr W last updated on 29/Dec/21

	Commented by Mathematification last updated on 29/Dec/21

	$P l e a s e s i r, w h i c h t e x t b o o k c a n o v e r s h a d o w t h i s ? ?$
	Commented by peter frank last updated on 29/Dec/21

	$thank you$
	Commented by mr W last updated on 29/Dec/21

	$t h i s i s a b o u t t h e a p p l i c a t i o n o f$ $a n a l y t i c g e o m e t r y . i h a v e n o$ $s p e c i a l t e x t b o o k t o r e c o m m e n d .$
	Commented by Mathematification last updated on 29/Dec/21

	$O k a y s i r . T h a n k y o u s o m u c h . I$ $h a v e m o r e q u e s t i o n s t o p o s t c o n c e r n i n g$ $i t .$
	Commented by mr W last updated on 29/Dec/21

	$a s s u m e t h e x - c o o r d i n a t e i s p, t h e n$ $t h e y - c o o r d i n a t e i s p^{2} . p i s u n k n o w n .$
	Commented by Mathematification last updated on 30/Dec/21

	$I s i t a r u l e t h a t t h e p o i n t o f i n t e r s e c t i o n$ $o f a p a r a b o l a a n d a c i r c l e i s a l w a y s i n t h e$ $f o r m (p, p^{2}) ?$ $I f n o, t h e n w h y {d i d n}^{'} t y o u a s s u m e t h a t i t$ $i s (p, 2 p) .$ $S i r ?$
	Commented by mr W last updated on 30/Dec/21

	$t h e e q n . o f t h e p a r a b o l a i s g i v e n a s$ $y = x^{2}, t h e r e f o r e e a c h p o i n t o n t h e$ $p a r a b o l a i s i n t h e f o r m (x, x^{2}) . i f x = p,$ $t h e n (p, p^{2}) .$
	Commented by Mathematification last updated on 30/Dec/21

	$T h a n k y o u s i r$
	Commented by Mathematification last updated on 30/Dec/21

	$P l e a s e s i r, I n e e d y o u r w h a t s a p p n u m b e r,$ $I w a n t t o b e m o r e i n t i m a t e w i t h y o u .$
	Commented by mr W last updated on 30/Dec/21

	$s o r r y s i r! b u t i {d o n}^{'} t w a n t t h a t . p l e a s e$ $u s e t h i s f o r u m f o r e x c h a n g e! i {d o n}^{'} t$ $a c c e p t p r i v a t e m e s s a g e s .$
	Commented by Mathematification last updated on 30/Dec/21

	$O k a y s i r$
	Commented by mr W last updated on 30/Dec/21

	$t h a n k s f o r y o u r u n d e r s t a n d i n g!$
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