Question Number 227055 by Spillover last updated on 28/Dec/25
$${A}\:{parabolic}\:{refector}\:{is}\:{formed}\:{by} \\ $$$${revolving}\:{the}\:{arc}\:{of}\:{the}\:{parabala} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{ax}\:\:{from}\:{x}=\mathrm{0}\:\:\:\:{to}\:\:{x}={h} \\ $$$${about}\:{the}\:{axis}.{If}\:{the}\:\:{diameter} \\ $$$${of}\:{the}\:{reflector}\:{is}\:\mathrm{2}{l}.{Show}\:{that} \\ $$$${the}\:{area}\:{of}\:{the}\:{reflecting}\:{surface}\:{is} \\ $$$$\frac{\pi{l}}{\mathrm{6}{h}^{\mathrm{2}} }\left\{\left({l}^{\mathrm{2}} +\mathrm{4}{h}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −{l}^{\mathrm{3}} \right\} \\ $$$$ \\ $$
Answered by Kassista last updated on 30/Dec/25
![y^2 =(√(4ax))=2(√(ax)) when x=h: y=2(√(ah))=radius=l y^′ =2.(1/( 2(√(ax)))).a=(a/( (√(ax))))=((√a)/( (√x))) Surface area is given by: A=2π∫_a ^( b) f(x).(√(1+f^′ (x)^2 ))dx ∴ A=2π∫_0 ^( h) 2(√(ax)).(√(1+(a/x)))dx = 4π(√a)∫_0 ^( h) (√x)(√((x+a)/x))dx =4π(√a)∫_0 ^( h) (x+a)^(1/2) dx= 4π(√a).(2/3)[(x+a)^(3/2) ]_0 ^h A= ((8(√a)π)/3)[(h+a)^(3/2) −a^(3/2) ] Rewriting a in terms of l and h: 2(√(ah))=l ⇒(√a)=(l/(2(√h))) ⇔a=(l^2 /(4h)) ∴ A=((8π)/3).(l/( 2(√h)))[(h+(l^2 /(4h)))^(1.5) −((l^2 /(4h)))^(1.5) ] = ((4πl)/(3(√h))).[(((4h^2 +l^2 )^(1.5) −l^3 )/((4h)^(1.5) ))] A=((4πl)/(3h^(0.5) )).(((4h^2 +l^2 )^(1.5) −l^3 )/(4^(1.5) .h^(1.5) )) ∴ A = ((πl)/(6h^2 ))[(4h^2 +l^2 )^(3/2) −l^3 ]](https://www.tinkutara.com/question/Q227097.png)
$${y}^{\mathrm{2}} =\sqrt{\mathrm{4}{ax}}=\mathrm{2}\sqrt{{ax}} \\ $$$${when}\:{x}={h}:\:{y}=\mathrm{2}\sqrt{{ah}}={radius}={l} \\ $$$${y}^{'} =\mathrm{2}.\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{ax}}}.{a}=\frac{{a}}{\:\sqrt{{ax}}}=\frac{\sqrt{{a}}}{\:\sqrt{{x}}} \\ $$$$ \\ $$$${Sur}\mathrm{f}{ace}\:{area}\:{is}\:{given}\:{by}:\:{A}=\mathrm{2}\pi\int_{{a}} ^{\:{b}} {f}\left({x}\right).\sqrt{\mathrm{1}+{f}^{'} \left({x}\right)^{\mathrm{2}} }{dx} \\ $$$$ \\ $$$$\therefore\:{A}=\mathrm{2}\pi\int_{\mathrm{0}} ^{\:{h}} \mathrm{2}\sqrt{{ax}}.\sqrt{\mathrm{1}+\frac{{a}}{{x}}}{dx}\:=\:\mathrm{4}\pi\sqrt{{a}}\int_{\mathrm{0}} ^{\:{h}} \sqrt{{x}}\sqrt{\frac{{x}+{a}}{{x}}}{dx} \\ $$$$ \\ $$$$=\mathrm{4}\pi\sqrt{{a}}\int_{\mathrm{0}} ^{\:{h}} \left({x}+{a}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dx}=\:\mathrm{4}\pi\sqrt{{a}}.\frac{\mathrm{2}}{\mathrm{3}}\left[\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{{h}} \: \\ $$$${A}=\:\frac{\mathrm{8}\sqrt{{a}}\pi}{\mathrm{3}}\left[\left({h}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −{a}^{\frac{\mathrm{3}}{\mathrm{2}}} \right] \\ $$$${Rewriting}\:{a}\:{in}\:{terms}\:{of}\:{l}\:{and}\:{h}: \\ $$$$\mathrm{2}\sqrt{{ah}}={l}\:\Rightarrow\sqrt{{a}}=\frac{{l}}{\mathrm{2}\sqrt{{h}}}\:\Leftrightarrow{a}=\frac{{l}^{\mathrm{2}} }{\mathrm{4}{h}} \\ $$$$ \\ $$$$\therefore\:{A}=\frac{\mathrm{8}\pi}{\mathrm{3}}.\frac{{l}}{\:\mathrm{2}\sqrt{{h}}}\left[\left({h}+\frac{{l}^{\mathrm{2}} }{\mathrm{4}{h}}\right)^{\mathrm{1}.\mathrm{5}} −\left(\frac{{l}^{\mathrm{2}} }{\mathrm{4}{h}}\right)^{\mathrm{1}.\mathrm{5}} \right]\:=\:\frac{\mathrm{4}\pi{l}}{\mathrm{3}\sqrt{{h}}}.\left[\frac{\left(\mathrm{4}{h}^{\mathrm{2}} +{l}^{\mathrm{2}} \right)^{\mathrm{1}.\mathrm{5}} −{l}^{\mathrm{3}} }{\left(\mathrm{4}{h}\right)^{\mathrm{1}.\mathrm{5}} }\right] \\ $$$${A}=\frac{\mathrm{4}\pi{l}}{\mathrm{3}{h}^{\mathrm{0}.\mathrm{5}} }.\frac{\left(\mathrm{4}{h}^{\mathrm{2}} +{l}^{\mathrm{2}} \right)^{\mathrm{1}.\mathrm{5}} −{l}^{\mathrm{3}} }{\mathrm{4}^{\mathrm{1}.\mathrm{5}} .{h}^{\mathrm{1}.\mathrm{5}} }\:\therefore\:{A}\:=\:\frac{\pi{l}}{\mathrm{6}{h}^{\mathrm{2}} }\left[\left(\mathrm{4}{h}^{\mathrm{2}} +{l}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −{l}^{\mathrm{3}} \right] \\ $$$$ \\ $$