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InterviewSolution
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Find the equation of the hyperbola whose vertices are `(+-7,0)` and the eccentricity is `(4)/(3).` |
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Answer» Since the vertices of the given hyperbola are of the form `(+-a,0)`,it is a horizontal hyperbola . Let the requied equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1.` then , its vertices are `(+-a,0)`. But , the vertices are `(+-7,0)` `therefore a=7hArra^(2)=49.` Also ,`e=( c)/(a)hArrc=ae=(7xx(4)/(3))=(28)/(3).` Now , `C^(2)=(a^(2)+b^(2))hArrb^(2)=(c^(2)-a^(2))=[((28)/(3))^(2)-49]=(343)/(9).` Thus`,a^(2)=49 and b^(2)=(343)/(9).` ` therefore` the required equation is `(x^(2))/(49)-(y^(2))/((343//9))=1hArr(x^(2))/(49)-(9y^(2))/(343)=1hArr 7x^(2)-9y^(2)=343.` |
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