JEE Advance - Mathematics (2021 - Paper 1 Online - No. 9)
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 __________.
$${L_1}:x\sqrt 2 + y - 1 = 0$$ 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 __________.
Answer
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)
$$ \Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270$$
$$ \Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270$$
$$ \Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9$$
$$C:\left| {{{x\sqrt 2 + y - 1} \over {\sqrt 3 }}} \right|\left| {{{x\sqrt 2 - y + 1} \over {\sqrt 3 }}} \right| = {\lambda ^2}$$
$$ \Rightarrow C:{{\left| {{{(x\sqrt 2 )}^2} - {{(y - 1)}^2}} \right|} \over {\sqrt 3 \times \sqrt 3 }} = {\lambda ^2}$$
$$ \Rightarrow C:\left| {2{x^2} - {{(y - 1)}^2}} \right| = 3{\lambda ^2}$$
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)
$$ \Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270$$
$$ \Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270$$
$$ \Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9$$
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