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 \underline\hspace2cm .

Straight Lines and Pair of Straight Lines PYQ — JEE Advanced 2021 · Shift 1 · Q28 with Solution

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Correct answer: 9

Given lines and locus

The lines are $L_1:\,\sqrt2 x+y-1=0, \qquad L_2:\,\sqrt2 x-y+1=0.$

For a point $P(x,y)$ , its distances from $L_1$ and $L_2$ are $d_1=\frac{|\sqrt2 x+y-1|}{\sqrt{2+1}}=\frac{|\sqrt2 x+y-1|}{\sqrt3},$ $d_2=\frac{|\sqrt2 x-y+1|}{\sqrt3}.$

Given that their product is $\lambda^2$ , $d_1d_2=\lambda^2.$ So, $\frac{|\sqrt2 x+y-1||\sqrt2 x-y+1|}{3}=\lambda^2.$

Squaring is not necessary because the product is already nonnegative; the locus is $|(\sqrt2 x+y-1)(\sqrt2 x-y+1)|=3\lambda^2.$

Now,

=(\sqrt2 x)^2-(y-1)^2 =2x^2-(y-1)^2.$$ Thus the locus $C$ is $$|2x^2-(y-1)^2|=3\lambda^2.$$ So $C$ consists of the pair of hyperbolas $$2x^2-(y-1)^2=\pm 3\lambda^2.$$ --- 2. **Intersect with the line $y=2x+1$** Substitute $y=2x+1$ into $$|2x^2-(y-1)^2|=3\lambda^2.$$ Since $y-1=2x$, $$|2x^2-(2x)^2|=|2x^2-4x^2|=|-2x^2|=2x^2.$$ Hence, $$2x^2=3\lambda^2 \quad\Rightarrow\quad x^2=\frac{3\lambda^2}{2}.$$ So the two intersection points are at $$x=\pm \sqrt{\frac{3\lambda^2}{2}}.$$ Their corresponding $y$-coordinates are $y=2x+1$. Let these points be $R$ and $S$. Since both lie on the line of slope $2$, the distance between them is $$RS=\sqrt{(\Delta x)^2+(\Delta y)^2}.$$ Now, $$\Delta x=2\sqrt{\frac{3\lambda^2}{2}}, \qquad \Delta y=2\Delta x=4\sqrt{\frac{3\lambda^2}{2}}.$$ Therefore, $$RS^2=(\Delta x)^2+(\Delta y)^2 =\left(2\sqrt{\frac{3\lambda^2}{2}}\right)^2+\left(4\sqrt{\frac{3\lambda^2}{2}}\right)^2.$$ Compute: $$RS^2=4\cdot \frac{3\lambda^2}{2}+16\cdot \frac{3\lambda^2}{2} =20\cdot \frac{3\lambda^2}{2}=30\lambda^2.$$ Given $$RS=\sqrt{270},$$ so $$RS^2=270=30\lambda^2.$$ Hence, $$\lambda^2=\frac{270}{30}=9.$$ --- 3. **Comparison with stored answer** Our derived value is $$\boxed{9}.$$ This matches the stored correct answer.

More from Straight Lines and Pair of Straight Lines

View written solutionFree

Correct answer: 9

Given lines and locus

The lines are $L_1:\,\sqrt2 x+y-1=0, \qquad L_2:\,\sqrt2 x-y+1=0.$

For a point $P(x,y)$ , its distances from $L_1$ and $L_2$ are $d_1=\frac{|\sqrt2 x+y-1|}{\sqrt{2+1}}=\frac{|\sqrt2 x+y-1|}{\sqrt3},$ $d_2=\frac{|\sqrt2 x-y+1|}{\sqrt3}.$

Given that their product is $\lambda^2$ , $d_1d_2=\lambda^2.$ So, $\frac{|\sqrt2 x+y-1||\sqrt2 x-y+1|}{3}=\lambda^2.$

Squaring is not necessary because the product is already nonnegative; the locus is $|(\sqrt2 x+y-1)(\sqrt2 x-y+1)|=3\lambda^2.$

Now,

=(\sqrt2 x)^2-(y-1)^2 =2x^2-(y-1)^2.$$ Thus the locus $C$ is $$|2x^2-(y-1)^2|=3\lambda^2.$$ So $C$ consists of the pair of hyperbolas $$2x^2-(y-1)^2=\pm 3\lambda^2.$$ --- 2. **Intersect with the line $y=2x+1$** Substitute $y=2x+1$ into $$|2x^2-(y-1)^2|=3\lambda^2.$$ Since $y-1=2x$, $$|2x^2-(2x)^2|=|2x^2-4x^2|=|-2x^2|=2x^2.$$ Hence, $$2x^2=3\lambda^2 \quad\Rightarrow\quad x^2=\frac{3\lambda^2}{2}.$$ So the two intersection points are at $$x=\pm \sqrt{\frac{3\lambda^2}{2}}.$$ Their corresponding $y$-coordinates are $y=2x+1$. Let these points be $R$ and $S$. Since both lie on the line of slope $2$, the distance between them is $$RS=\sqrt{(\Delta x)^2+(\Delta y)^2}.$$ Now, $$\Delta x=2\sqrt{\frac{3\lambda^2}{2}}, \qquad \Delta y=2\Delta x=4\sqrt{\frac{3\lambda^2}{2}}.$$ Therefore, $$RS^2=(\Delta x)^2+(\Delta y)^2 =\left(2\sqrt{\frac{3\lambda^2}{2}}\right)^2+\left(4\sqrt{\frac{3\lambda^2}{2}}\right)^2.$$ Compute: $$RS^2=4\cdot \frac{3\lambda^2}{2}+16\cdot \frac{3\lambda^2}{2} =20\cdot \frac{3\lambda^2}{2}=30\lambda^2.$$ Given $$RS=\sqrt{270},$$ so $$RS^2=270=30\lambda^2.$$ Hence, $$\lambda^2=\frac{270}{30}=9.$$ --- 3. **Comparison with stored answer** Our derived value is $$\boxed{9}.$$ This matches the stored correct answer.

Straight Lines and Pair of Straight Lines question

Straight Lines and Pair of Straight Lines question

More from Straight Lines and Pair of Straight Lines

Straight Lines and Pair of Straight Lines question

Straight Lines and Pair of Straight Lines question

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