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Find the equation of the hyperbola where foci are `(0,+-12)`and the length of the latus rectum is 36. |
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Answer» The focus of the hyperbola lies on y-axis, terefore, let the equation of the hyperbola is `(y^(2))/(b^(2))-(x^(2))/(a^(2))=1` Co-ordinates of foci `=(0,pmbe)` be=12 and latus rectum `(2a^(2))/(b)=36` `rArr" "a^(2)=18b` `rArr" "b^(2)(e^(2)-1)=18b` `rArr" "144-b^(2)=18b` `rArr""b^(2)+18b-144=0` `rArr" "(b+24)(b-6)=0` `rArr" "b=-24orb=6` b=-24 (which is not possible) `:.` b=6 Therefore, equation of hyerbola `(y^(2))/(36)-(x^(2))/(108)=1` `rArr" "3y^(2)-x^(2)=108`. |
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