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Find the equation of the hyperbola whose eccentricity is `(5)/(3)` and whose vertices are `(0,+-6).` Also, find the coordinates of its foci . |
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Answer» Since the vertices of the given hyperbola are of the form `(-0,+-a)`, it is a vertical hyperbola . Let the required equation be `(y^(2))/(a^(2))-(x^(2))/(b^(2))=1.` Then , its vertices are `(0,+-a)` But , it is given that the vertices are `(0,+-6).` `therefore a=6.Also , e=(5)/(3).` Now , `(c )/(a)=eimpliesc=ae=(6xx(5)/(3))=10.` `And ,c^(2)=(a^(2)+b^(2))hArr b^(2)=(c^(2)-a^(2))={(10)^(2)-6^(2)}=(100-36)=64.` THus , `a^(2)=6^(2)=36 and b^(2)=64.` Hence , the required equation is `(y^(2))/(36)-(x^(2))/(64)=1.` Also , the coordinates of its foci are `(0,+-c),i.e., (0,+-10).` |
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