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Find the vertices, eccentricity foci, rquation of directrices and length of latus rectum of the hyperbola `9x^(2)-25y^(2)=225`. |
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Answer» Equation of hyperbola. `9x^(2)-25y^(2)=225` `rArr" "(x^(2))/(25)-(y^(2))/(9)=1` Here `a^(2)=25andb^(2)=9` `rArr" "a=5andb=3` Co-ordinates of vertices `=(pma,0)=(pm5,0)`. Eccentricity e `=sqrt(1+(b^(2))/(a^(2)))=sqrt(1+(9)/(25))=sqrt(34)/(5)`. Co-ordinates of foci `=(pmae,0)=(pm5xx(sqrt(34))/(5),0)=(pmsqrt(34),0)`. Equation of directrices `x=pm(a)/(e)` `rArr""x=pm(5)/(sqrt(34)/(5))` `rArr""x=pm(5)/(sqrt(34))`. Length of latus rectum `=(2b^(2))/(a)=(2xx9)/(5)=(18)/(5)`. |
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