Question Number 51269 by peter frank last updated on 25/Dec/18
$${Find}\:{the}\:{ecentricity}\:{If} \\ $$$$\left(\mathrm{1}\right){lactus}\:{rectum}\:{is}\:{half} \\ $$$${major}\:{axis} \\ $$$$\left(\mathrm{2}\right){lactus}\:{rectum}\:{is}\:{half} \\ $$$${minor}\:{axis} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18
$$\left.\mathrm{1}\right){e}=\frac{{c}}{{a}} \\ $$$${LT}\:\:\:\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}}={a}\:\:\:\:\mathrm{2}{b}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$ \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:\:\rightarrow{for}\:{elipse} \\ $$$$\:\:\:={a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\:\:\:\:{c}=\pm\frac{{a}}{\:\sqrt{\mathrm{2}}}\:\:\: \\ $$$${so}\:{e}=\frac{{c}}{{a}}=\pm\frac{{a}}{\:\sqrt{\mathrm{2}}\:×{a}}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${for}\:{hyperbola} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${c}=\pm\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:\:{a} \\ $$$${e}=\frac{{c}}{{a}}=\pm\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\: \\ $$$$\left.\mathrm{2}\right)\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}}={b} \\ $$$${a}=\mathrm{2}{b} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:{for}\:{ellipse} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${e}=\frac{{c}}{{a}}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${for}\:{hyperbola} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${c}^{\mathrm{2}} =\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${e}=\frac{{c}}{{a}}=\pm\frac{\sqrt{\mathrm{5}}\:}{\mathrm{2}} \\ $$$$ \\ $$
Commented by peter frank last updated on 25/Dec/18
$${sir}\:{the}\:{ans}\:{given}\:{are} \\ $$$$\left(\mathrm{1}\right){e}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{e}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18
$${ok}\:{i}\:{have}\:{igonred}\:\pm\:{sign}…{let}\:{me}\:{add}\:\pm\:{sign} \\ $$