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
$${rH}=\frac{{R}\left(\mathrm{2}{r}+\mathrm{2}\sqrt{{r}^{\mathrm{2}} +{H}^{\mathrm{2}} }\right)}{\mathrm{2}}={rR}+{R}\sqrt{{r}^{\mathrm{2}} +{H}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}^{\mathrm{2}} \left({H}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \left({r}^{\mathrm{2}} +{H}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{r}^{\mathrm{2}} \left({H}^{\mathrm{2}} \right)+{r}^{\mathrm{2}} {R}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} {HR}={r}^{\mathrm{2}} {R}^{\mathrm{2}} +{R}^{\mathrm{2}} {H}^{\mathrm{2}} \\ $$$$\Rightarrow{H}^{\mathrm{2}} \left({r}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{2}{r}^{\mathrm{2}} {HR} \\ $$$$\Rightarrow\frac{{r}^{\mathrm{2}} −{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} {R}^{\mathrm{2}} }=\frac{\mathrm{2}}{{HR}}\Rightarrow\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{1}}{{r}^{\mathrm{2}} }=\frac{\mathrm{2}}{{HR}} \\ $$
Answered by mr W last updated on 18/May/24
$$\frac{{R}}{{r}}=\frac{\sqrt{\left({H}−{R}\right)^{\mathrm{2}} −{R}^{\mathrm{2}} }}{{H}}=\frac{\sqrt{{H}\left({H}−\mathrm{2}{R}\right)}}{{H}} \\ $$$$\frac{{R}^{\mathrm{2}} }{{r}^{\mathrm{2}} }=\mathrm{1}−\frac{\mathrm{2}{R}}{{H}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{R}^{\mathrm{2}} }−\frac{\mathrm{1}}{{r}^{\mathrm{2}} }=\frac{\mathrm{2}}{{RH}} \\ $$