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If the tangent at `(1,1)`on `y^2=x(2-x)^2`meets the curve again at `P ,`then find coordinates of `Pdot`A. (4, 4)B. (2, 0)C. (9/4, 3/8)D. `(3,3^(1//2))` |
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Answer» Correct Answer - C `2y(dy)/(dx)=(2-x)^(2)-2(2-x)` `"So "(dy)/(dx):|_("(1,1)")``=-(1)/(2)` Therefore, the equation of tangent at (1,1) is `y-1=-(1)/(2)(x-1)` `rArr" "y=(-x+3)/(2)` The intersection of the tangent and the curve is given by `(1//4)(-x+3)^(2)=x(4+x^(2)-4x)` `rArr" "4x^(3)-17x^(2)+22x-9=0` `rArr" "(x-1)(4x^(2)-13x+9)=0` `rArr" "(x-1)(4x^(2)-13x+9)=0` `rArr" "(x-1)^(2)(4x-9)=0` Since x = 1 is already the point of tangency, `x=9//4` and `y^(2)=(9)/(2)(2-(9)/(4))^(2)=(9)/(64)`. Thus, the required point is `(9//4, 3//8)`. |
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