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1

### JEE Advanced 2021 Paper 1 Online

Numerical
Consider the lines L1 and L2 defined by

$${L_1}:x\sqrt 2 + y - 1 = 0$$ and $${L_2}:x\sqrt 2 - y + 1 = 0$$

For a fixed constant $$\lambda$$, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is $$\lambda$$2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is $$\sqrt {270}$$. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.

The value of D is __________.
Your Input ________

## Answer

Correct Answer is 77.14

## Explanation

According to the question,

$$C:\left| {{{x\sqrt 2 + y - 1} \over {\sqrt 3 }}} \right|\left| {{{x\sqrt 2 - y + 1} \over {\sqrt 3 }}} \right| = {\lambda ^2}$$

$$\Rightarrow C:{{\left| {{{(x\sqrt 2 )}^2} - {{(y - 1)}^2}} \right|} \over {\sqrt 3 \times \sqrt 3 }} = {\lambda ^2}$$

$$\Rightarrow C:\left| {2{x^2} - {{(y - 1)}^2}} \right| = 3{\lambda ^2}$$

Let R $$\equiv$$ (x1, y1) and S(x2, y2)

$$\because$$ C cuts y $$-$$ 1 = 2x at R and S.

So, $$\left| {2{x^2} - 4{x^2}} \right| = 3{\lambda ^2}$$

$$\Rightarrow x = \pm \sqrt {{3 \over 2}} \left| \lambda \right|$$

$$\therefore$$ $$\left| {{x_1} - {x_2}} \right| = \sqrt 6 \left| \lambda \right|$$

and $$\left| {{y_1} - {y_2}} \right| = 2\left| {{x_1} - {x_2}} \right| = 2\sqrt 6 \left| \lambda \right|$$

$$\because$$ RS2 = 270 (given)

$$\Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270$$

$$\Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270$$

$$\Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9$$

Now, mid-point of RS is $$\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right) \equiv (0,1)$$ and slope of RS = 2 and slope of $$R'S' = {{ - 1} \over 2}$$

$$\therefore$$ Equation of $$R'S':y - 1 = - {1 \over 2}x$$ i.e. 2y $$-$$ 2 = $$-$$x

$$\Rightarrow x + 2y - 2 = 0$$

On solving x + 2y $$-$$ 2 = 0 with C, we get

$${x^2} = {{12} \over 7}{\lambda ^2} \Rightarrow \left| {{x_1} - {x_2}} \right| = 2\sqrt {{{12} \over 7}} \left| \lambda \right|$$

and $$\left| {{y_1} - {y_2}} \right| = {1 \over 2}\left| {{x_1} - {x_2}} \right| = \sqrt {{{12} \over 7}} \left| \lambda \right|$$

Hence, $$D \equiv {(R'S')^2} = {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2}$$

$$= {{12} \over 7} \times 9 \times 5 = {{12 \times 45} \over 7} \approx 77.14$$
2

### JEE Advanced 2021 Paper 1 Online

Numerical
Consider the lines L1 and L2 defined by

$${L_1}:x\sqrt 2 + y - 1 = 0$$ and $${L_2}:x\sqrt 2 - y + 1 = 0$$

For a fixed constant $$\lambda$$, let C be the locus of a point P such that the product of the distance of P from L1 and the distance of P from L2 is $$\lambda$$2. The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is $$\sqrt {270}$$. Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.

The value of $$\lambda$$2 is __________.
Your Input ________

## Answer

Correct Answer is 9

## Explanation

According to the question,

$$C:\left| {{{x\sqrt 2 + y - 1} \over {\sqrt 3 }}} \right|\left| {{{x\sqrt 2 - y + 1} \over {\sqrt 3 }}} \right| = {\lambda ^2}$$

$$\Rightarrow C:{{\left| {{{(x\sqrt 2 )}^2} - {{(y - 1)}^2}} \right|} \over {\sqrt 3 \times \sqrt 3 }} = {\lambda ^2}$$

$$\Rightarrow C:\left| {2{x^2} - {{(y - 1)}^2}} \right| = 3{\lambda ^2}$$

Let R $$\equiv$$ (x1, y1) and S(x2, y2)

$$\because$$ C cuts y $$-$$ 1 = 2x at R and S.

So, $$\left| {2{x^2} - 4{x^2}} \right| = 3{\lambda ^2}$$

$$\Rightarrow x = \pm \sqrt {{3 \over 2}} \left| \lambda \right|$$

$$\therefore$$ $$\left| {{x_1} - {x_2}} \right| = \sqrt 6 \left| \lambda \right|$$

and $$\left| {{y_1} - {y_2}} \right| = 2\left| {{x_1} - {x_2}} \right| = 2\sqrt 6 \left| \lambda \right|$$

$$\because$$ RS2 = 270 (given)

$$\Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270$$

$$\Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270$$

$$\Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9$$
3

### JEE Advanced 2014 Paper 1 Offline

Numerical
For a point $$P$$ in the plane, Let $${d_1}\left( P \right)$$ and $${d_2}\left( P \right)$$ be the distance of the point $$P$$ from the lines $$x - y = 0$$ and $$x + y = 0$$ respectively. The area of the region $$R$$ consisting of all points $$P$$ lying in the first quadrant of the plane and satisfying $$2 \le {d_1}\left( P \right) + {d_2}\left( P \right) \le 4$$, is
Your Input ________

## Answer

Correct Answer is 6

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