Question-72773

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 .

Facebook Tweet Pin