Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

Subjective

Find the angle subtended by the double ordinate of length 2a of the parabola y^{2} = ax at its vertex.

.

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(m_{1}) = $${{0 - a} \over {0 - a}} = 1$$

Slope of OB(m_{2}) = $${{0 - a} \over {0 + a}} = -1$$

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

Subjective

Prove that for all values of m, except zero the st. line $$y = mx + {a \over m}$$ touches the parabola y^{2} = 4ax

.

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 y^{2} = 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 y^{2} = 4ax.

3

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.

.

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

On those following papers in Subjective

Number in Brackets after Paper Indicates No. of Questions

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Logarithms

Sequence and Series

Quadratic Equations

Complex Numbers

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Matrices and Determinants

Vector Algebra

Three Dimensional Geometry

Probability

Statistics

Sets and Relations

Functions

Definite Integration

Application of Integration

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Differential Equations

Straight Lines and Pair of Straight Lines

Circle

Parabola

Ellipse and Hyperbola