The equation of the hyperbola with vertices at (0,`pm`6) and eccentricity 5/3 is….. And its foci are …….

The equation of the hyperbola with vertices at (0,`pm`6) and eccentric

1.	The equation of the hyperbola with vertices at (0,`pm`6) and eccentricity 5/3 is….. And its foci are …….
Answer» Let the equation of the hyperbola be -`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` Then vertices=(0,`pm`b)=(0,`pm`6) `therefore b=6 and e=5//3` `becausee=sqrt(1+(a^(2))/(b^(2)))rArr25/9=1+(a^(2))/36` `rArr (25-9)/9=(a^(2))/36rArr16=(a^(2))/4rArra^(2)=48` So, the equation of hyperbola is `(-x^(2))/48+(y^(2))/36=1rArr(y^(2))/36-(x^(2))/48=1` `because` Foci=(0,`pm`be)=(0,`pm5/3xx6`)=(0,`pm`10)

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The equation of the hyperbola with vertices at (0,`pm`6) and eccentricity 5/3 is….. And its foci are …….