InterviewSolution
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InterviewSolution
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If the latus rectum of a hyperbola forms an equilateral triangle with the vertex at the center of the hyperbola ,then find the eccentricity of the hyperbola. |
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Answer» `(a^2e^2)/(a^2)-y^2/b^2=1` `e^2-1=y^2/b^2` `y^2=b^2(e^2-1)` `y=pmbsqrt(e^2-1)` `tan30^o=(AD)/(OD)` `1/sqrt3=(bsqrt(e^2-1))/(ae)` Squaring both sides `1/3=(b^2(e^2-1))/(a^2e^2)` `e^2/(e^2-1)=3(b^2/a^2)` `e=c/a=sqrt(a^2+b^2)/a^2=sqrt(1+b^2/a^2` `e^2=1+b^2/a^2` `b^2/a^2=e^2-1` `e^2=3(e^2-1)` `e^2/3=(e^2-1)^2` `pme/sqrt3=e^2-1` `sqrt3e^2-e-sqrt3=0` `e=(1pmsqrt(1+12))/(2sqrt3)` `e=(1+sqrt13)/(2sqrt3)` `e=(-1+sqrt13)/(2sqrt3)` |
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