Question Number 52058 by peter frank last updated on 02/Jan/19
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jan/19
$${eqn}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\: \\ $$$${focus}\left({ae},\mathrm{0}\right) \\ $$$${shortest}\:{distance}={a}−{ae} \\ $$$${farthest}\:{distance}={a}+{ae} \\ $$$${a}−{ae}=\mathrm{3}×\mathrm{10}^{\mathrm{6}} \:{km} \\ $$$${a}−{a}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{3}×\mathrm{10}^{\mathrm{6}} {km} \\ $$$${a}=\frac{\mathrm{3}×\mathrm{3}×\mathrm{10}^{\mathrm{6}} }{\mathrm{2}}=\mathrm{4}.\mathrm{5}×\mathrm{10}^{\mathrm{6}} \\ $$$${farthest}={a}+{ae} \\ $$$$=\mathrm{4}.\mathrm{5}×\mathrm{10}^{\mathrm{6}} +\mathrm{4}.\mathrm{5}×\mathrm{10}^{\mathrm{6}} ×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\mathrm{6}×\mathrm{10}^{\mathrm{6}} {km}... \\ $$$${pld}\:{check}... \\ $$
Commented by peter frank last updated on 02/Jan/19
$${correct}\:{sir}.{thank}\:{you} \\ $$