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Question: If the tangent at (1,1) on \(y^{2} = x(2 - x)^{2}\) meets the curves again at P, then P is...

If the tangent at (1,1) on y2=x(2−x)2y^{2} = x(2 - x)^{2} meets the curves again at P, then P is

A

(4,4)

B

(-1,2)

C

(94,38)\left( \frac{9}{4},\frac{3}{8} \right)

D

None

Answer

(94,38)\left( \frac{9}{4},\frac{3}{8} \right)

Explanation

Solution

2ydydx=(2−x)2−2x(2−x),\mathbf{2y}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\left( \mathbf{2 - x} \right)^{\mathbf{2}}\mathbf{- 2x}\left( \mathbf{2 - x} \right)\mathbf{,} so  dydx∣(1,1)=12[1−2]=−12\left. \ \frac{dy}{dx} \right|_{(1,1)} = \frac{1}{2}\lbrack 1 - 2\rbrack = - \frac{1}{2}

Therefore, the equation of tangent at (1,1) is

y−1=−12(x−1)y - 1 = - \frac{1}{2}(x - 1)

⇒ 2y−2=−x+12y - 2 = - x + 1

⇒ y=−x+32y = \frac{- x + 3}{2}

The intersection of the tangent and the curve is given by

(14)(−x+3)2=x(4+x2−4x)\left( \frac{1}{4} \right)( - x + 3)^{2} = x\left( 4 + x^{2} - 4x \right)

⇒ x2−6x+9=16x+4x3−16x2x^{2} - 6x + 9 = 16x + 4x^{3} - 16x^{2}

⇒ 4x3−17x2+22x−9=04x^{3} - 17x^{2} + 22x - 9 = 0

⇒ (x−1)(4x2−13x+9)=0(x - 1)\left( 4x^{2} - 13x + 9 \right) = 0

⇒ (x−1)2(4x−9)=0(x - 1)^{2}(4x - 9) = 0

Since xx =1 is already the point of tangency x=94x = \frac{9}{4}

and y2=94(2−94)2=964y^{2} = \frac{9}{4}\left( 2 - \frac{9}{4} \right)^{2} = \frac{9}{64}

Thus the required point is (94,38)\left( \frac{9}{4},\frac{3}{8} \right)