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Question Number 207543 by efronzo1 last updated on 18/May/24

Commented by efronzo1 last updated on 18/May/24

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Answered by A5T last updated on 18/May/24

rH=((R(2r+2(√(r^2 +H^2 ))))/2)=rR+R(√(r^2 +H^2 )) ⇒r^2 (H−R)^2 =R^2 (r^2 +H^2 ) ⇒r^2 (H^2 )+r^2 R^2 −2r^2 HR=r^2 R^2 +R^2 H^2 ⇒H^2 (r^2 −R^2 )=2r^2 HR ⇒((r^2 −R^2 )/(r^2 R^2 ))=(2/(HR))⇒(1/R^2 )−(1/r^2 )=(2/(HR))

$${rH}=\frac{{R}\left(\mathrm{2}{r}+\mathrm{2}\sqrt{{r}^{\mathrm{2}} +{H}^{\mathrm{2}} }\right)}{\mathrm{2}}={rR}+{R}\sqrt{{r}^{\mathrm{2}} +{H}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}^{\mathrm{2}} \left({H}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \left({r}^{\mathrm{2}} +{H}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{r}^{\mathrm{2}} \left({H}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} {R}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} {HR}={r}^{\mathrm{2}} {R}^{\mathrm{2}} +{R}^{\mathrm{2}} {H}^{\mathrm{2}} \\ $$$$\Rightarrow{H}^{\mathrm{2}} \left({r}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{2}{r}^{\mathrm{2}} {HR} \\ $$$$\Rightarrow\frac{{r}^{\mathrm{2}} −{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} {R}^{\mathrm{2}} }=\frac{\mathrm{2}}{{HR}}\Rightarrow\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{1}}{{r}^{\mathrm{2}} }=\frac{\mathrm{2}}{{HR}} \\ $$

Answered by mr W last updated on 18/May/24

(R/r)=((√((H−R)^2 −R^2 ))/H)=((√(H(H−2R)))/H) (R^2 /r^2 )=1−((2R)/H) ⇒(1/R^2 )−(1/r^2 )=(2/(RH))

$$\frac{{R}}{{r}}=\frac{\sqrt{\left({H}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }}{{H}}=\frac{\sqrt{{H}\left({H}−\mathrm{2}{R}\right)}}{{H}} \\ $$$$\frac{{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{2}{R}}{{H}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{1}}{{r}^{\mathrm{2}} }=\frac{\mathrm{2}}{{RH}} \\ $$

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