NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

WB JEE 2009

Subjective

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

Answer

.

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

Answer

.

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.

Answer

.

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.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12