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Question Number 72773 by ajfour last updated on 02/Nov/19

Commented by ajfour last updated on 02/Nov/19

$I f p a r a b o l a i n x z p l a n e h a s e q .$ $z = {a x}^{2} f i n d e q . o f s h a d o w o f$ $t h i s p a r a b o l a o n g r o u n d (x y p l a n e) .$
Commented by MJS last updated on 02/Nov/19

$you get a hyperbola with$ $y = \frac{d}{2} (\sqrt{1 + \frac{4 {a x}^{2}}{h}} - 1)$
	Answered by mr W last updated on 02/Nov/19

	${l e t}^{'} s s a y t h e s h a d o w o f p o i n t P (x, z) i n$ $p l a n e x z i s p o i n t Q (u, v) i n p l a n e x y .$ $i n 3 D - s y s t e m :$ $S (0, - d, h)$ $P (x, 0, z)$ $Q (u, v, 0)$ $\frac{u}{x} = \frac{v + d}{d} = \frac{- h}{z - h}$ $\Rightarrow x = \frac{d u}{v + d}$ $\Rightarrow z = \frac{h v}{v + d}$ $e x a m p l e :$ $t h e s h a d o w o f p a r a b o l a z = {a x}^{2} :$ $z = {a x}^{2}$ $\Rightarrow \frac{h v}{v + d} = a {(\frac{d u}{v + d})}^{2}$ $\Rightarrow v (v + d) = \frac{{a d}^{2} u^{2}}{h}$ $o r$ $\Rightarrow y (y + d) = \frac{{a d}^{2} x^{2}}{h} \leftarrow e q n . o f s h a d o w$ $o r$ $\Rightarrow y = (\sqrt{1 + \frac{4 {a x}^{2}}{h}} - 1) \frac{d}{2}$ $t h e s h a d o w o f c i r c l e x^{2} + {(z - r)}^{2} = r^{2} :$ $x^{2} + {(z - r)}^{2} = r^{2}$ ${(\frac{d u}{v + d})}^{2} + {(\frac{h v}{v + d} - r)}^{2} = r^{2}$ $o r$ $\Rightarrow d^{2} x^{2} = h y [(2 r - h) y + 2 r d]$ $o r$ $\Rightarrow y = \frac{- r h d + d \sqrt{r^{2} h^{2} - (2 r - h) {h x}^{2}}}{2 r - h}$
	Commented by ajfour last updated on 02/Nov/19

	$T h a n k y o u M j S S i r, m r W S i r .$
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