Consider the lines L_ 1 and L_ 2 defined by L_ 1 : x 2 + y-1=0 and L_ 2 : x 2 - y+1=0 For a fixed constant λ , let C be the locus of a point P such that the product of the distance of P from L_ 1 and the distance of P

Correct answer: 0 Let the point P is (h, k) Distance of P from L _ 1 = | h 2 + k - 1 ( 2 ) ^ 2 + 1 ^ 2 | = | h 2 + k - 1 3 | Distance of P from L _ 2 = | h 2 - k + 1 ( 2 ) ^ 2 + 1 ^ 2 | = | h 2 - k + 1 3 | ∴ The equation of the locus of P is | h 2 + k - 1 3 | × | h 2 - k + 1 3 | = λ ^ 2 |( h 2 + k - 1 3 )( h 2 - k + 1 3 )| = λ ^ 2 ⇒ |2 h ^ 2 - ( k - 1 ) ^ 2 | = 3 λ ^ 2 Hence, the equation of the locus is |2 x ^ 2 - ( y - 1 ) ^ 2 | = 3 λ ^ 2 The line is y = 2x + 1 or y - 1 = 2x By substituting the value of y in the equation of the curve C , we get |2 x ^ 2 - ( y - 1 ) ^ 2 | = 3 λ ^ 2 ⇒ |2 x ^ 2 - ( 2x ) ^ 2 | = 3 λ ^ 2 ⇒ 2 x ^ 2 = 3 λ ^ 2 ⇒ x = ± (3)/(2) λ ⇒ x _ 2 - x _ 1 = | 6 λ | Also, y - 1 = 2x ⇒ y _ 2 - 1 = 2 x _ 2 and y _ 1 - 1 = 2 x _ 1 ⇒ y _ 2 - y _ 1 = 2( x _ 2 - x _ 1 ) ⇒ y _ 2 - y _ 1 = | 2 6 λ | Given RS = 270 ⇒ ( x _ 2 - x _ 1 ) ^ 2 + ( y _ 2 - y _ 1 ) ^ 2 = 270 ⇒ ( 6 λ ) ^ 2 + ( 2 6 λ ) ^ 2 = 270 30 λ ^ 2 = 270 λ ^ 2 = 9

JEE Advanced 2021Paper 1

Exam: JEE Advanced
Year: 2021
Paper: Paper 1
Subject: Mathematics
Chapter: Straight Lines

JEE Advanced 2021 · Paper 1 · Mathematics · Straight Lines

Consider the lines L_ 1 and L_ 2 defined by L_ 1 : x 2 + y-1=0 and...

Consider the lines

L_{1}

and

L_{2}

defined by
\\ L_{1}: x

\sqrt{2}+

y-1=0

\text{ 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

L_{1}

and the distance of

P

from

L_{2}

\lambda^{2}

. The line

y=2 x+1

meets

C

at two points

R

and

S

, where the distance between

R

and

S

\sqrt{270}

. \\ Let the perpendicular bisector of

R S

meet

C

at two distinct points

R^{\prime}

and

S^{\prime}

. Let

D

be the square of the distance between

R^{\prime}

and

S^{\prime}

The value of ${λ}^{2}$ is

Correct answer

Step-by-step solution

Let the point $P$ is \left(h, k\right) $Distance of$ P $from$ {L}_{1} = \left|\dfrac{h\sqrt{2} + k - 1}{\sqrt{( \sqrt{2}{)}^{2} + {1}^{2}}}\right|

$= \left|\dfrac{h\sqrt{2} + k - 1}{\sqrt{3}}\right|$

Distance of $P$ from ${L}_{2} = \left|\dfrac{h\sqrt{2} - k + 1}{\sqrt{( \sqrt{2}{)}^{2} + {1}^{2}}}\right|$ = \left|\dfrac{h\sqrt{2} - k + 1}{\sqrt{3}}\right| $$ ∴ $The equation of the locus of$ P $is\left|\dfrac{h\sqrt{2} + k - 1}{\sqrt{3}}\right| \times \left|\dfrac{h\sqrt{2} - k + 1}{\sqrt{3}}\right| = {\lambda}^{2}$

\left|\left(\dfrac{h\sqrt{2} + k - 1}{\sqrt{3}}\right)\left(\dfrac{h\sqrt{2} - k + 1}{\sqrt{3}}\right)\right| = {\lambda}^{2} $$ ⇒ \left|2{h}^{2} - ( k - 1{)}^{2}\right| = 3{\lambda}^{2} $Hence, the equation of the locus is \left|2{x}^{2} - ( y - 1{)}^{2}\right| = 3{\lambda}^{2}$

The line is $y = 2x + 1 or y - 1 = 2x$

By substituting the value of $y$ in the equation of the curve $C$ , we get

\left|2{x}^{2} - ( y - 1{)}^{2}\right| = 3{\lambda}^{2} $$ ⇒ \left|2{x}^{2} - ( 2x{)}^{2}\right| = 3{\lambda}^{2} $$ ⇒ 2{x}^{2} = 3{\lambda}^{2} $$ ⇒ x = \pm \sqrt{\dfrac{3}{2}}\lambda $$ ⇒ {x}_{2} - {x}_{1} = | \sqrt{6}\lambda | $Also,$ y - 1 = 2x $$ ⇒ {y}_{2} - 1 = 2{x}_{2} and {y}_{1} - 1 = 2{x}_{1} $$ ⇒ {y}_{2} - {y}_{1} = 2\left({x}_{2} - {x}_{1}\right) $$ ⇒ {y}_{2} - {y}_{1} = | 2\sqrt{6}\lambda | $Given$ RS = \sqrt{270} $$ ⇒ \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}} = \sqrt{270} $$ ⇒ ( \sqrt{6}\lambda{)}^{2} + ( 2\sqrt{6}\lambda{)}^{2} = 270 $$ 30{\lambda}^{2} = 270 $$ {\lambda}^{2} = 9$

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