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1

### JEE Advanced 2021 Paper 2 Online

Numerical
Let E be the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$. For any three distinct points P, Q and Q' on E, let M(P, Q) be the mid-point of the line segment joining P and Q, and M(P, Q') be the mid-point of the line segment joining P and Q'. Then the maximum possible value of the distance between M(P, Q) and M(P, Q'), as P, Q and Q' vary on E, is _______.

## Explanation

As we know that, in a triangle, sides joining the mid-points of two sides is half and parallel to the third side.

$$\therefore$$ $${M_1}{M_2} = {1 \over 2}QQ'$$

Maximum value of QQ' is AA'

Hence, maximum value of $${M_1}{M_2} = {1 \over 2}AA' = 4$$
2

### JEE Advanced 2017 Paper 1 Offline

Numerical
For how many values of p, the circle x2 + y2 + 2x + 4y $$-$$ p = 0 and the coordinate axes have exactly three common points?

## Explanation

We will consider three cases

Case 1 : Circle passes through origin, that is, p = 0 the equation of circle becomes

x2 + y2 + 2x + 4y = 0

x = 0 $$\Rightarrow$$ y2 + 4y = 0 $$\Rightarrow$$ y(y + 4) = 0

y = 0, $$-$$4

y = 0 $$\Rightarrow$$ x2 + 2x = 0 $$\Rightarrow$$ x(x + 2) = 0 $$\Rightarrow$$ x = 0, $$-$$ 2

Case 2 : Circle touches y axis then circle will intersect x axis at two distinct points

Put y = 0 in equation of circle, we get

x2 + 2x $$-$$ p = 0

Now from g2 $$-$$ c > 0 and f2 $$-$$ c = 0

$$\Rightarrow$$ 12 $$-$$ ($$-$$p) > 0 and 22 $$-$$ ($$-$$p) = 0

1 + p > 0 and 4 + p = 0

p > $$-$$1 and p = $$-$$ 4

which is a contradiction. Also, for p = $$-$$4 we get

x2 $$-$$ 2x + 4 = 0

$$x = {{2 \pm \sqrt {4 - 4 \times 4} } \over 2} = {{2 \pm 2\sqrt 3 i} \over 2} = 1 \pm \sqrt 3 i$$

They are imaginary roots.

Therefore, Case 2 is not possible.

Case 3 : Circle touches x axis. Then the circle will intersect y axis at two distinct points.

Substituting x = 0, we get

y2 + 4y $$-$$ p = 0

Now, from g2 $$-$$ C = 0 and f2 $$-$$ C > 0, we have

12 $$-$$ ($$-$$p) = 0 and 22 = ($$-$$p) > 0

p = $$-$$1 and 4 + p > 0 $$\Rightarrow$$ p > $$-$$4

Therefore, the equation of circle becomes

y2 + 4y + 1 = 0

$$\Rightarrow y = {{ - 4 \pm \sqrt {16 - 4} } \over 2} = {{ - 4 \pm \sqrt {12} } \over 2}$$

$$y = {{ - 4 \pm \sqrt {4 \times 3} } \over 2}$$

$$= {{ - 4 \pm 2\sqrt 3 } \over 2} = - 2 \pm \sqrt 3$$ real values

Thus, Case 3 is possible.

Thus, for p = 0 and p = $$-$$1 the circle and coordinates have exactly three points in common. Hence, the correct answer is 2.

3

### JEE Advanced 2015 Paper 2 Offline

Numerical
Suppose that the foci of the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$ are $$\left( {{f_1},0} \right)$$ and $$\left( {{f_2},0} \right)$$ where $${{f_1} > 0}$$ and $${{f_2} < 0}$$. Let $${P_1}$$ and $${P_2}$$ be two parabolas with a common vertex at $$(0,0)$$ and with foci at $$\left( {{f_1},0} \right)$$ and $$\left( 2{{f_2},0} \right)$$, respectively. Let $${T_1}$$ be a tangent to $${P_1}$$ which passes through $$\left( 2{{f_2},0} \right)$$ and $${T_2}$$ be a tangent to $${P_2}$$ which passes through $$\left( {{f_1},0} \right)$$. If $${m_1}$$ is the slope of $${T_1}$$ and $${m_2}$$ is the slope of $${T_2}$$, then the value of $$\left( {{1 \over {m_1^2}} + m_2^2} \right)$$ is

## Explanation

$${e^2} = 1 - {{{b^2}} \over {{a^2}}} = 1 - {5 \over 9} = {4 \over 9}$$

The foci are ($$\pm$$ ae, 0) i.e. (2, 0) and ($$-$$2, 0).

The parabola P1 is $${y^2} = 8x$$ and P2 is $${y^2} = - 16x$$

As tangent with slope m1 to P1 passes through ($$-$$4, 0), we have

$$y = {m_1}x + {2 \over {{m_1}}}$$ giving $$0 = - 4{m_1} + {2 \over {{m_1}}}$$

i.e. $$4m_1^2 = 2 \Rightarrow m_1^2 = {1 \over 2}$$

Again for tangent with slope m2 to P2 passing through (2, 0), we have

$$y = {m_2}x - {4 \over {{m_2}}} \Rightarrow 0 = 2{m_2} - {4 \over {{m_2}}}$$

$$\Rightarrow 2m_2^2 = 4$$ $$\therefore$$ $$m_2^2 = 2$$

Thus, $${1 \over {m_1^2}} + m_2^2 = 2 + 2 = 4$$

4

### JEE Advanced 2015 Paper 1 Offline

Numerical
Let the curve $$C$$ be the mirror image of the parabola $${y^2} = 4x$$ with respect to the line $$x+y+4=0$$. If $$A$$ and $$B$$ are the points of intersection of $$C$$ with the line $$y=-5$$, then the distance between $$A$$ and $$B$$ is

## Explanation

Let, P(t2, 2t) be any point on the parabola y2 = 4x. C be the mirror image of the parabola y2 = 4x with respect to the line UV : x + y + 4 = 0.

The curve C cuts the line KL : y = $$-$$5 at A and B.

Let, B($$\alpha$$, $$\beta$$) be the image of the point P(t2, 2t).

Clearly, PB $$\bot$$ UV and PQ = QB.

$$\therefore$$ $${{\alpha - {t^2}} \over {\beta - 2t}} \times ( - 1) = - 1$$

or, $$\alpha - {t^2} = \beta - 2t$$ ...... (1)

The point of intersection of the lines UV and KL is R.

Let us join P and R.

From $$\Delta$$PQR and $$\Delta$$BQR,

(i) BQ = PQ [$$\because$$ B is the image of P]

(ii) $$\angle$$PQR = $$\angle$$RQB = 90$$^\circ$$ [$$\because$$ PB $$\bot$$ UV]

(iii) QR common

$$\therefore$$ $$\Delta$$PQR $$\cong$$ $$\Delta$$BQR [by SAS congruence criterion]

$$\therefore$$ $$\angle$$QRP = $$\angle$$BRQ [CPCT]

$$\because$$ slope of x + y + 4 = 0 is $$-$$1,

$$\therefore$$ $$\angle$$UTO = 135$$^\circ$$

$$\therefore$$ $$\angle$$OTR = 45$$^\circ$$

Again, X'X || KL and UV transversal.

$$\therefore$$ $$\angle$$OTR = $$\angle$$TRB = 45$$^\circ$$ $$\therefore$$ $$\angle$$BRQ = $$\angle$$QRP = 45$$^\circ$$

$$\therefore$$ $$\angle$$PRB = 90$$^\circ$$ $$\therefore$$ PR $$\bot$$ KL

$$\therefore$$ coordinates of R are (t2, $$\beta$$).

$$\because$$ the point R lies on KL,

$$\therefore$$ $$\beta$$ = $$-$$5

Again, the point R lies on the straight line x + y + 4 = 0.

$$\therefore$$ t2 + $$\beta$$ + 4 = 0

or, t2 $$-$$ 5 + 4 = 0 [$$\because$$ $$\beta$$ = $$-$$5]

or, t2 = 1 or, t = $$\pm$$ 1

when t = 1, $$\beta$$ = $$-$$5, then (1) $$\Rightarrow$$ $$\alpha$$ $$-$$ 1 = $$-$$ 5 $$-$$ 2 or $$\alpha$$ = $$-$$6

when, t = $$-$$1, $$\beta$$ = $$-$$5, then (1) $$\Rightarrow$$ $$\alpha$$ $$-$$ 1 = $$-$$ 5 + 2 or, $$\alpha$$ = $$-$$2

So, the coordinates of A and B are ($$-$$6, $$-$$5) and ($$-$$2, $$-$$5) respectively.

$$\therefore$$ AB = 4 units

So, the distance between A and B is 4 units.

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