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Answer» `"orIm"((x+iy)^(2))=3` `"or"2xy=3` which is a RECTANGULAR hyperbola having eccentricity `sqrt2`.  `TAN30^(@)=(b^(2)//a)/(2ae)` `"or"(2)/(sqrt3)e=e^(2)-1` `"or"sqrt3e^(2)-2e-sqrt3=0` LTBRGT `"or"e=(2pmsqrt(4+12))/(2sqrt3)=(2pm4)/(2sqrt3)` `"or"e=(3)/(sqrt3)=sqrt3` (c) Eccentricity of hyperbola`=(AB)/(PA-PB)=(6)/(4)=(3)/(2)` If the eccentricity of conjugate hyperbola is e', then `(1)/((3//2)^(2))+(1)/(e'^(2))=1` `"or"e'=(3)/(sqrt5)` (d) Angle between the asymptotes is `tan^(-1)|(2ab)/(a^(2)-b^(2))|=(pi)/(3)` `"or"|(2xxa//b)/((a^(2)//b^(2)))|=sqrt3` `"or"(2sqrt(e'^(2)-1))/(|e'^(2)-2|)=sqrt3` where e' is the eccentricity of conjugate hyperbola. THEREFORE, `e'=2.` |
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