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1

### JEE Main 2021 (Online) 27th August Morning Shift

Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :
A
3x2 $$-$$ 2y $$-$$ 6 = 0
B
3x2 + 2y $$-$$ 6 = 0
C
2x2 + 3y $$-$$ 9 = 0
D
2x2 $$-$$ 3y + 9 = 0

## Explanation

A(0, 6) and B(2t, 0)

Perpendicular bisector of AB is

$$(y - 3) = {t \over 3}(x - t)$$

So, $$C = \left( {0,3 - {{{t^2}} \over 3}} \right)$$

Let P be (h, k)

$$h = {t \over 2};k = \left( {3 - {{{t^2}} \over 6}} \right)$$

$$\Rightarrow k = 3 - {{4{h^2}} \over 6} \Rightarrow 2{x^2} + 3y - 9 = 0$$
2

### JEE Main 2021 (Online) 26th August Morning Shift

Let ABC be a triangle with A($$-$$3, 1) and $$\angle$$ACB = $$\theta$$, 0 < $$\theta$$ < $${\pi \over 2}$$. If the equation of the median through B is 2x + y $$-$$ 3 = 0 and the equation of angle bisector of C is 7x $$-$$ 4y $$-$$ 1 = 0, then tan$$\theta$$ is equal to :
A
$${1 \over 2}$$
B
$${3 \over 4}$$
C
$${4 \over 3}$$
D
2

## Explanation

$$\therefore$$ $$M\left( {{{a - 3} \over 2},{{b + 1} \over 2}} \right)$$ lies on 2x + y $$-$$ 3 = 0

$$\Rightarrow$$ 2a + b = 11 ...........(i)

$$\because$$ C lies on 7x $$-$$ 4y = 1

$$\Rightarrow$$ 7a $$-$$ 4b = 1 ......... (ii)

$$\therefore$$ by (i) and (ii) : a = 3, b = 5

$$\Rightarrow$$ C(3, 5)

$$\therefore$$ mAC = 2/3

Also, mCD = 7/4

$$\Rightarrow \tan {\theta \over 2} = \left| {{{{2 \over 3} - {4 \over 4}} \over {1 + {{14} \over {12}}}}} \right| \Rightarrow \tan {\theta \over 2} = {1 \over 2}$$

$$\Rightarrow \tan \theta = {{2.{1 \over 2}} \over {1 - {1 \over 4}}} = {4 \over 3}$$
3

### JEE Main 2021 (Online) 25th July Evening Shift

Let the equation of the pair of lines, y = px and y = qx, can be written as (y $$-$$ px) (y $$-$$ qx) = 0. Then the equation of the pair of the angle bisectors of the lines x2 $$-$$ 4xy $$-$$ 5y2 = 0 is :
A
x2 $$-$$ 3xy + y2 = 0
B
x2 + 4xy $$-$$ y2 = 0
C
x2 + 3xy $$-$$ y2 = 0
D
x2 $$-$$ 3xy $$-$$ y2 = 0

## Explanation

Equation of angle bisector of homogeneous
equation of pair of straight line ax2 + 2hxy + by2 is

$${{{x^2} - {y^2}} \over {a - b}} = {{xy} \over h}$$

for x2 – 4xy – 5y2 = 0

a = 1, h = – 2, b = – 5

So, equation of angle bisector is

$${{{x^2} - {y^2}} \over {1 - ( - 5)}} = {{xy} \over { - 2}}$$

$${{{x^2} - {y^2}} \over 6} = {{xy} \over { - 2}}$$

$$\Rightarrow {x^2} - {y^2} = - 3xy$$

So, combined equation of angle bisector is $${x^2} + 3xy - {y^2} = 0$$
4

### JEE Main 2021 (Online) 27th July Evening Shift

Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals of the parallelogram is 11x + 7y = 9, then other diagonal passes through the point :
A
(1, 2)
B
(2, 2)
C
(2, 1)
D
(1, 3)

## Explanation

Both the lines pass through origin.

point D is equal to intersection of 4x + 5y = 0 & 11x + 7y = 9

So, coordinates of point $$D = \left( {{5 \over 3}, - {4 \over 3}} \right)$$

Also, point B is point of intersection of 7x + 2y = 0 and 11x + 7y = 9

So, coordinates of point $$B = \left( { - {2 \over 3},{7 \over 3}} \right)$$

diagonals of parallelogram intersect at middle let middle point of B, D

$$\Rightarrow \left( {{{{5 \over 3} - {2 \over 3}} \over 2},{{{{ - 4} \over 3} + {7 \over 3}} \over 2}} \right) = \left( {{1 \over 2},{1 \over 2}} \right)$$

equation of diagonal AC

$$\Rightarrow (y - 0) = {{{1 \over \alpha } - 0} \over {{1 \over \alpha } - 0}}(x - 0)$$

$$y = x$$

diagonal AC passes through (2, 2)

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