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Question Number 196460 by peter frank last updated on 25/Aug/23

Commented by mr W last updated on 25/Aug/23

$q u e s t i o n i s n o t c l e a r .$ $t_{1}, t_{2}, t_{3} a r e t h r e e p o i n t s . w h a t d o y o u$ $m e a n w i t h t_{1} + t_{2} + t_{3} ?$
Commented by universe last updated on 25/Aug/23

$i t h i n k t_{1} t_{2} t_{3} a r e s l o p e s o f n o r m a l s$
Commented by Spillover last updated on 26/Aug/23
Check you solution https://m.facebook.com/story.php?story_fbid=pfbid0hkqsvcktd24vmL7bxE8SHoaJdmVahXFrgpdvX3XSog1UD4cQCC5oNeqdbkLwQ1kKl&id=100084816875916&mibextid=Nif5oz
Commented by mr W last updated on 26/Aug/23

$h o w c o u l d a q u e s t i o n b e a n s w e r e d$ $w h e n i t i s e v e n n o t c l e a r w h a t t h e$ $q u e s t i o n i s ?$
Commented by peter frank last updated on 26/Aug/23

$thank you spillover$
	Answered by mr W last updated on 27/Aug/23

	Commented by mr W last updated on 27/Aug/23

	$s a y t h e e q u a t i o n o f a n o r m a l f r o m$ $p o i n t P (h, k) i s$ $y = k + t (x - h)$ $s a y t h e n o r m a l m e e t s t h e p a r a b o l a$ $a t p o i n t A (u, v) .$ $\tan θ = \frac{d x}{d y} ∣_{y = v} = \frac{v}{2 a} = - t$ $\Rightarrow v = - 2 a t$ $u = \frac{v^{2}}{4 a} = \frac{{(- 2 a t)}^{2}}{4 a} = {a t}^{2}$ $v = k + t (u - h)$ $- 2 a t = k + t ({a t}^{2} - h)$ ${a t}^{3} + (2 a - h) t + k = 0$ $t h r e e r o o t s c o u l d e x i s t f o r t, t h a t$ $m e a n s t h r e e n o r m a l s t h r o u g h$ $p o i n t (h, k) c o u l d e x i s t^{)} .$ $t_{1} + t_{2} + t_{3} = \frac{0}{a} = 0 ✓$ $^{)} t h e c o n d i t i o n t h a t t h r e e n o r m a l s$ $e x i s t i s t h a t t h e e q u a t i o n h a s t h r e e$ $r e a l r o o t s,$ $\Rightarrow {(- \frac{k}{2 a})}^{2} + {(\frac{2 a - h}{3 a})}^{3} < 0$ $\Rightarrow \frac{{a k}^{2}}{4} + \frac{{(2 a - h)}^{3}}{27} < 0$ $t h a t m e a n s p o i n t P (h, k) m u s t l i e$ $i n s i d e t h e g r e e n a r e a :$ $(e x a m p l e f o r a = 0.5)$
	Commented by mr W last updated on 27/Aug/23

	Commented by peter frank last updated on 08/Sep/23

	$thank you mr W .$
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