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Find the vertices, eccentricity, foci, equation of directrices length of latus rectum for the hyperbola `3y^(2)-x^(2)=27`. |
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Answer» `3y^(2)-x^(2)=27`. `rArr" "(y^(2))/(9)-(x^(2))/(27)-1` which is the equation of conjugate hyperbola Here `b^(2)=9anda^(2)=27` `rArr""b=3anda=3sqrt(3)` Co-ordinates of vertices `(0,pmb)=(0,pm3)`. Eccentricity e `=sqrt(1+(a^(2))/(b^(2)))` `=sqrt(1+(27)/(9))=2`. Co-ordinates of foci `=(0,pmbe)` `=(0,pm3xx2)` `(0,pm6)`. Equation of directrices `y-pm(b)/(e)" "rArr" "y=pm(3)/(2)`. Length of latus rectum `=(2a^(2))/(b)=(2xx27)/(3)=18`. |
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