If the tangent at $(1, 1)$ on $y^2 = x(2 - x)^2$ meets the curve again at $P$ then $P$ is (A) $(4, 4)$ (B) $(-1, 2)$ (C) $(3, 6)$ (D) $\left(\dfrac94, \dfrac38\right)$

If the tangent at (1, 1) on y² = x(2 - x)² meets t – Class 12 Mathematics

Miscellaneous Exercise 2(I) · Q108

(1, 1)

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Differentiating $y^2=x(2-x)^2$ implicitly: $2y y'=(2-x)^2+x\cdot2(2-x)(-1)=(2-x)[(2-x)-2x]=(2-x)(2-3x)$ .

At $(1,1)$ : $y'=\dfrac{(2-1)(2-3)}{2(1)}=\dfrac{(1)(-1)}{2}=-\dfrac12$ .

Tangent: $y-1=-\dfrac12(x-1) \Rightarrow 2y-2=-(x-1) \Rightarrow x+2y-3=0 \Rightarrow y=\dfrac{3-x}{2}$ .

Substituting into the curve: $\left(\dfrac{3-x}{2}\right)^2=x(2-x)^2 \Rightarrow (3-x)^2=4x(2-x)^2$ .

Expanding: $9-6x+x^2 = 4x(4-4x+x^2)=16x-16x^2+4x^3$ , so $4x^3-17x^2+22x-9=0$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set A1 markMCQ
$y = x^2 + 4x + 1$
(a) $x + 10y = 8$
(b) $10x + y = 8$
(c) $10x - y = 8$
(d) $x - 10y = 8$
›Reveal solutionSolution

The tangent at $x=3$ is $10x - y = 8$ .

Given $y = x^2+4x+1$ . The slope is

$\frac{dy}{dx} = 2x+4.$

At $x=3$ : slope $= 2(3)+4 = 10$ , and $y = 9+12+1 = 22$ , so the point is $(3,22)$ .

Equation of the tangent: …

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CBSE 2023Set ANNUAL1 markMCQ
$y^2 = 4ax$
(a) $t$
(b) $\dfrac{2}{t}$
(c) $\dfrac{1}{t}$
(d) None of these
›Reveal solutionSolution

Use parametric differentiation: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ .

For $y^2=4ax$ , the point $(at^2,2at)$ satisfies $x=at^2$ , $y=2at$ .

$\dfrac{dx}{dt}=2at$ , $\dfrac{dy}{dt}=2a$

…

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CBSE 2019Set ANNUAL1 mark
$y = x^3 - x$
›Reveal solutionSolution

The slope of the tangent at a point equals $\dfrac{dy}{dx}$ evaluated at that point.

Given $y = x^3 - x$

$\dfrac{dy}{dx} = 3x^2 - 1$

…

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CBSE 2018Set ANNUAL1 markMCQ
$x = a\cos^3\theta$
(a) $1$
(b) $2$
(c) $-1$
(d) None of these
›Reveal solutionSolution

Find $dy/dx$ for the parametric curve and evaluate at $\theta=\pi/4$ .

$x = a\cos^3\theta \Rightarrow \dfrac{dx}{d\theta} = -3a\cos^2\theta\sin\theta$ .

$y = a\sin^3\theta \Rightarrow \dfrac{dy}{d\theta} = 3a\sin^2\theta\cos\theta$ .

…

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CBSE 2017Set ANNUAL1 markMCQ
$y = x^3 - 11x + 5$
(a) $(-2, 0)$
(b) $(3, 7)$
(c) $(0, 2)$
(d) $(2, -9)$
›Reveal solutionSolution

Matching the curve's slope to the tangent's slope $1$ gives the point $(2,-9)$ .

$y=x^3-11x+5\Rightarrow \dfrac{dy}{dx}=3x^2-11$ . The tangent line $y=x-11$ has slope $1$ , so $3x^2-11=1\Rightarrow x^2=4\Rightarrow x=\pm2$ .

…

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Q.If the tangent at $(1, 1)$ on $y^2 = x(2 - x)^2$ meets the curve again at $P$ then $P$ is (A) $(4, 4)$ (B) $(-1, 2)$ (C) $(3, 6)$ (D) $\left(\dfrac94, \dfrac38\right)$

Concept understanding — Tangent and Normal to a Curve

Q.If the tangent at (1,1)(1, 1)(1,1) on y2=x(2−x)2y^2 = x(2 - x)^2y2=x(2−x)2 meets the curve again at PPP then PPP is (A) (4,4)(4, 4)(4,4) (B) (−1,2)(-1, 2)(−1,2) (C) (3,6)(3, 6)(3,6) (D) (94,38)\left(\dfrac94, \dfrac38\right)(49​,83​)

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Q.If the tangent at $(1, 1)$ on $y^2 = x(2 - x)^2$ meets the curve again at $P$ then $P$ is (A) $(4, 4)$ (B) $(-1, 2)$ (C) $(3, 6)$ (D) $\left(\dfrac94, \dfrac38\right)$

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