Equation of Tangent at a Point (x,y) in Terms of f'(x)
If the tangen...
Question
If the tangent at (1, 1) on y2=x(2−x)2 meets the curve again at P, then P is
A
(4, 4)
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B
(-1,2)
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C
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D
None of these
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Solution
The correct option is C 2y dydx=(2−x)2 - 2x(2 - x), so dydx|(1,1)=12[1−2]=−12 Therefore, the equation of tangent at (1, 1) is y - 1 = -12(x−1) ⇒ 2y - 2 = -x + 1 ⇒ y = −x+32 The intersection of the tangent and the curve is given by (14)(−x+3)2=x(4+x2−4x) ⇒x2−6x+9=16x+4x3−16x2 ⇒4x3−17x2+22x−9=0 ⇒(x−1)(4x2−13x+9)=0 ⇒(x−1)2(4x−9)=0 Since x = 1 is already the point of tangency x = 94 and y2 = 94(2−94)2=964 Thus the required point is (94,38)