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Find-the-ecentricity-If-1-lactus-rectum-is-half-major-axis-2-lactus-rectum-is-half-minor-axis-

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

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