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Find the equation of the hyperbola, the length ofwhose latusrectum is 8 and eccentricity is `3//sqrt(5)dot`A. `5x^(2)-4y^(2)=100`B. `4x^(2)-5y^(2)=100`C. `-4x^(2)+5y^(2)=100`D. `-5x^(2)+4y^(2)=100` |
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Answer» Let the equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` ……..`(i)` We have, Length of the latusrectum `=8` `implies(2b^(2))/(a)=8` `impliesb^(2)=4a` `impliesa^(2)(e^(2)-1)=4a` `impliesa(e^(2)-1)=4impliesa((9)/(5)-1)=4impliesa=5` Putting `a=5 "in" b^(2)=4a`, we get `b^(2)=20`. Hence, the equation of the required hyperbola is `(x^(2))/(25)-(y^(2))/(20)=1` |
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