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Question Number 199568 by ajfour last updated on 05/Nov/23

Commented by ajfour last updated on 05/Nov/23

$H e n c e w h a t i s m i f x = a = m b$
	Answered by ajfour last updated on 05/Nov/23
	$sin α=(a/(b−a)) x^2 +y^2 =b^2 A(bsin 2α, bcos 2α) centre of X circle (p,p) & radius r Q[(b−a)cos α, a] let y−bcos 2α=m(x−bsin 2α) (({a−bcos 2α−m[(b−a)cos α−bsin 2α]})/( (√(1+m^2 ))))=a even p−bcos 2α−m[p−bsin 2α]=r(√(1+m^2 )) so we found m<0 then p>0 r=p(√2)sin ((π/4)−2α)$
	$\sin α = \frac{a}{b - a}$ $x^{2} + y^{2} = b^{2}$ $A (b \sin 2 α, b \cos 2 α)$ $c e n t r e o f X c i r c l e (p, p) & r a d i u s r$ $Q [(b - a) \cos α, a]$ $l e t y - b \cos 2 α = m (x - b \sin 2 α)$ $\frac{{a - b \cos 2 α - m [(b - a) \cos α - b \sin 2 α]}}{\sqrt{1 + m^{2}}} = a$ $e v e n$ $p - b \cos 2 α - m [p - b \sin 2 α] = r \sqrt{1 + m^{2}}$ $s o w e f o u n d m < 0 t h e n p > 0$ $r = p \sqrt{2} \sin (\frac{π}{4} - 2 α)$
	Commented by ajfour last updated on 05/Nov/23
	https://youtu.be/QoHc39Vmp3E?si=WaxfcQ7XGn5uEq2b
	Commented by ajfour last updated on 05/Nov/23

	$m y l e c t u r e p a r a b o l a 3$
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