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### WB JEE 2008

Subjective

If the tangent to the parabola y = x(2 $$-$$ x) at the point (1, 1) intersects the parabola at P. Find the co-ordinate of P.

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## Explanation

$$\therefore$$ $$y = x(2 - x)$$ ..... (i)

$$\therefore$$ $${{dy} \over {dx}} = {d \over {dx}}(2x - {x^2}) = 2 - 2x$$

Slope of tangent at (1, 1) is

$${\left( {{{dy} \over {dx}}} \right)_{(1,1)}} = 2 - 2 \times 1 = 0$$

$$\therefore$$ equation of tangent at (1, 1) is

$$y - {y_1} = {\left( {{{dy} \over {dx}}} \right)_{({x_1},{y_1})}}(x - {x_1})$$

$$\Rightarrow y - 1 = 0(x - 1) \Rightarrow y = 1$$ ..... (ii)

To find the co-ordinates of P, we solve (i) and (ii)

$$\therefore$$ $$2x - {x^2} = 1$$

$$\Rightarrow {x^2} - 2x + 1 = 0$$

$$\Rightarrow {(x - 1)^2} = 0$$

$$\Rightarrow$$ x = 1, 1

$$\because$$ both values of x are same.

$$\therefore$$ both points coincide. So, co-ordinates of P are (1, 1), i.e. given point (1, 1) is point P.

Note : If tangent is drawn at any point on parabola, it will never intersect the parabola again at other point.

So, given point (1, 1) is the point P.

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