Question Number 72773 by ajfour last updated on 02/Nov/19
Commented by ajfour last updated on 02/Nov/19
$${If}\:{parabola}\:{in}\:{xz}\:{plane}\:{has}\:{eq}. \\ $$$${z}={ax}^{\mathrm{2}} \:\:\:{find}\:{eq}.\:{of}\:{shadow}\:{of} \\ $$$${this}\:{parabola}\:{on}\:{ground}\:\left({xy}\:{plane}\right). \\ $$
Commented by MJS last updated on 02/Nov/19
$$\mathrm{you}\:\mathrm{get}\:\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{with} \\ $$$${y}=\frac{{d}}{\mathrm{2}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{4}{ax}^{\mathrm{2}} }{{h}}}−\mathrm{1}\right) \\ $$
Answered by mr W last updated on 02/Nov/19
![let′s say the shadow of point P(x,z) in plane xz is point Q(u,v) in plane xy. in 3D−system: S(0,−d,h) P(x,0,z) Q(u,v,0) (u/x)=((v+d)/d)=((−h)/(z−h)) ⇒x=(du/(v+d)) ⇒z=((hv)/(v+d)) example: the shadow of parabola z=ax^2 : z=ax^2 ⇒((hv)/(v+d))=a((du/(v+d)))^2 ⇒v(v+d)=((ad^2 u^2 )/h) or ⇒y(y+d)=((ad^2 x^2 )/h) ← eqn. of shadow or ⇒y=((√(1+((4ax^2 )/h)))−1)(d/2) the shadow of circle x^2 +(z−r)^2 =r^2 : x^2 +(z−r)^2 =r^2 ((du/(v+d)))^2 +(((hv)/(v+d))−r)^2 =r^2 or ⇒d^2 x^2 =hy[(2r−h)y+2rd] or ⇒y=((−rhd+d(√(r^2 h^2 −(2r−h)hx^2 )))/(2r−h))](https://www.tinkutara.com/question/Q72777.png)
$${let}'{s}\:{say}\:{the}\:{shadow}\:{of}\:{point}\:{P}\left({x},{z}\right)\:{in} \\ $$$${plane}\:{xz}\:{is}\:{point}\:{Q}\left({u},{v}\right)\:{in}\:{plane}\:{xy}. \\ $$$${in}\:\mathrm{3}{D}−{system}: \\ $$$${S}\left(\mathrm{0},−{d},{h}\right) \\ $$$${P}\left({x},\mathrm{0},{z}\right) \\ $$$${Q}\left({u},{v},\mathrm{0}\right) \\ $$$$\frac{{u}}{{x}}=\frac{{v}+{d}}{{d}}=\frac{−{h}}{{z}−{h}} \\ $$$$\Rightarrow{x}=\frac{{du}}{{v}+{d}} \\ $$$$\Rightarrow{z}=\frac{{hv}}{{v}+{d}} \\ $$$$ \\ $$$${example}: \\ $$$${the}\:{shadow}\:{of}\:{parabola}\:{z}={ax}^{\mathrm{2}} : \\ $$$${z}={ax}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{hv}}{{v}+{d}}={a}\left(\frac{{du}}{{v}+{d}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{v}\left({v}+{d}\right)=\frac{{ad}^{\mathrm{2}} {u}^{\mathrm{2}} }{{h}} \\ $$$${or} \\ $$$$\Rightarrow{y}\left({y}+{d}\right)=\frac{{ad}^{\mathrm{2}} {x}^{\mathrm{2}} }{{h}}\:\:\leftarrow\:{eqn}.\:{of}\:{shadow} \\ $$$${or} \\ $$$$\Rightarrow{y}=\left(\sqrt{\mathrm{1}+\frac{\mathrm{4}{ax}^{\mathrm{2}} }{{h}}}−\mathrm{1}\right)\frac{{d}}{\mathrm{2}} \\ $$$$ \\ $$$${the}\:{shadow}\:{of}\:{circle}\:{x}^{\mathrm{2}} +\left({z}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} : \\ $$$${x}^{\mathrm{2}} +\left({z}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left(\frac{{du}}{{v}+{d}}\right)^{\mathrm{2}} +\left(\frac{{hv}}{{v}+{d}}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${or} \\ $$$$\Rightarrow{d}^{\mathrm{2}} {x}^{\mathrm{2}} ={hy}\left[\left(\mathrm{2}{r}−{h}\right){y}+\mathrm{2}{rd}\right] \\ $$$${or} \\ $$$$\Rightarrow{y}=\frac{−{rhd}+{d}\sqrt{{r}^{\mathrm{2}} {h}^{\mathrm{2}} −\left(\mathrm{2}{r}−{h}\right){hx}^{\mathrm{2}} }}{\mathrm{2}{r}−{h}} \\ $$
Commented by ajfour last updated on 02/Nov/19
$${Thank}\:{you}\:{MjS}\:{Sir},\:{mrW}\:{Sir}. \\ $$