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The point `(3tan (theta +60^(@)),2 tan(theta +30^(@)))` lies on the conic, then its centre is `(theta` is the parameter)A. `(-3sqrt(3),2sqrt(3))`B. `(3sqrt(3),-2sqrt(3))`C. `(-3sqrt(3),-2sqrt(3))`D. (0,0) |
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Answer» Correct Answer - A Let `(3 tan (theta + 60^(@)),2 tan (theta + 30^(@)) -= (h,k)` `:. tan (theta + 60^(@)) = (h)/(3)` (1) and `tan (theta + 30^(@)) = (k)/(2)` (2) `tan 30^(@) = tan [(theta + 60^(@))- (theta + 30^(@))]` `rArr (1)/(sqrt(3)) = (tan (theta+60^(@))-tan(theta+30^(@)))/(1+tan (theta+60^(@))tan (theta+30^(@)))` `rArr (1)/(sqrt(3)) =((x)/(3)-(y)/(2))/(1+(xy)/(6))` `rArr xy - 2sqrt(3)x + 3sqrt(3)y + 6 =0` `rArr (x+3sqrt(3)) (y-2sqrt(3)) + 24 =0` `rArr` center is `(-3sqrt(2),2sqrt(3))` |
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