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Question-52058




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 (x^2 /a^2 )+(y^2 /b^2 )=1    focus(ae,0)  shortest distance=a−ae  farthest distance=a+ae  a−ae=3×10^6  km  a−a×(1/3)=3×10^6 km  a=((3×3×10^6 )/2)=4.5×10^6   farthest=a+ae  =4.5×10^6 +4.5×10^6 ×(1/3)  =6×10^6 km...  pld check...
$${eqn}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\: \\ $$$${focus}\left({ae},\mathrm{0}\right) \\ $$$${shortest}\:{distance}={a}−{ae} \\ $$$${farthest}\:{distance}={a}+{ae} \\ $$$${a}−{ae}=\mathrm{3}×\mathrm{10}^{\mathrm{6}} \:{km} \\ $$$${a}−{a}×\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{3}×\mathrm{10}^{\mathrm{6}} {km} \\ $$$${a}=\frac{\mathrm{3}×\mathrm{3}×\mathrm{10}^{\mathrm{6}} }{\mathrm{2}}=\mathrm{4}.\mathrm{5}×\mathrm{10}^{\mathrm{6}} \\ $$$${farthest}={a}+{ae} \\ $$$$=\mathrm{4}.\mathrm{5}×\mathrm{10}^{\mathrm{6}} +\mathrm{4}.\mathrm{5}×\mathrm{10}^{\mathrm{6}} ×\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\mathrm{6}×\mathrm{10}^{\mathrm{6}} {km}… \\ $$$${pld}\:{check}… \\ $$
Commented by peter frank last updated on 02/Jan/19
correct sir.thank you
$${correct}\:{sir}.{thank}\:{you} \\ $$

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