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1

### WB JEE 2009

Subjective

Find the angle subtended by the double ordinate of length 2a of the parabola y2 = ax at its vertex.

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## Explanation Let AB be the double ordinate and $$\angle$$AOB be the angle made by AB at the origin.

AB = 2a (given)

co-ordinates of A $$\equiv$$ (a, a)

co-ordinates of B $$\equiv$$ (a, $$-$$a)

co-ordinates of O $$\equiv$$ (0, 0)

Slope of OA(m1) = $${{0 - a} \over {0 - a}} = 1$$

Slope of OB(m2) = $${{0 - a} \over {0 + a}} = -1$$

Angle between OA and OB

$$\tan \theta = \left| {{{{m_2} - {m_1}} \over {1 + {m_1}{m_2}}}} \right| = \left| {{{ - 1 - 0} \over {1 + ( - 1)}}} \right| = {1 \over 0}$$ = Not defined

$$0 = {\tan ^{ - 1}}(\infty ) \Rightarrow \theta = \pi /2$$.

$$\angle$$AOB = $$\pi$$/2.

2

### WB JEE 2008

Subjective

Prove that for all values of m, except zero the st. line $$y = mx + {a \over m}$$ touches the parabola y2 = 4ax

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## Explanation

Find the point of intersection of

$$y = mx + {a \over m}$$ and $${y^2} = 4ax$$

$$\therefore$$ $$y = {{4a} \over m}\left( {y - {a \over m}} \right) \Rightarrow {y^2} - {{4a} \over m}y + {{4{a^2}} \over {{m^2}}} = 0$$

$$\Rightarrow {\left( {y - {{2a} \over m}} \right)^2} = 0 \Rightarrow y = {{2a} \over m},{{2a} \over m}$$

$$\because$$ both values of y are same, so there is only one point of intersection. But in case of parabola y2 = 4ax if line intersects at one point whose slope is not zero, then the line touches the parabola. So, $$y = mx + {a \over m}$$ touches the parabola y2 = 4ax.

3

### 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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