Consider the lines L_1 and L_2 defined by

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Question

$P(x, y) \quad\left|\frac{\sqrt{2} x+y-1}{\sqrt{3}}\right|\left|\frac{\sqrt{2} x-y+1}{\sqrt{3}}\right|=\lambda^2$

$\left|\frac{2 x^2-(y-1)^2}{3}\ri

Vedclass · Accepted Answer

$9,77.14$

Vedclass · Answer

$P(x, y) \quad\left|\frac{\sqrt{2} x+y-1}{\sqrt{3}}ight|\left|\frac{\sqrt{2} x-y+1}{\sqrt{3}}ight|=\lambda^2$

$\left|\frac{2 x^2-(y-1)^2}{3}ight|=\lambda^2, C :\left|2 x^2-(y-1)^2ight|=3 \lambda^2$

$	ext { line } y =2 x +1, RS =\sqrt{\left( x _1- x _2ight)^2+\left( y _1- y _2ight)^2}, R \left( x _1, y _1ight) 	ext { and } S\left( x _2, y _2ight)$

$y _1=2 x _1+1 	ext { and } y _2=2 x _2+1 \Rightarrow\left( y _1- y _2ight)=2\left( x _1- x _2ight)$

$	ext { RS }=\sqrt{5\left( x _1- x _2ight)^2}=\sqrt{5}\left| x _1- x _2ight|$

solve curve $C$ and line $y=2 x+1$ we get

$\left|2 x^2-(2 x)^2ight|=3 \lambda^2 \Rightarrow x^2=\frac{3 \lambda^2}{2}$

$	ext { RS }=\sqrt{5}\left|\frac{2 \sqrt{3} \lambda}{\sqrt{2}}ight|=\sqrt{30 \lambda}$$=\sqrt{270} \Rightarrow 30 \lambda^2=270 \Rightarrow \lambda^2=9$

$\perp$ bisector of $RS$

$T \equiv\left(\frac{ x _1+ x _2}{2}, \frac{y_1+y_2}{2}ight)$

Here $x _1+ x _2=0$

$T=(0,1)$

Equation of

$R^{\prime} S^{\prime}:(y-1)=-\frac{1}{2}(x-0) \Rightarrow x+2 y=2$

$R^{\prime}\left(a_1, b_1ight) S^{\prime}\left(a_2, b_2ight)$

$D=\left(a_1-a_2ight)^2+\left(b_1-b_2ight)^2=5\left(b_1-b_2ight)^2$

solve $x+2 y=2$ and $\left|2 x^2-(y-1)^2ight|=3 \lambda^2$

$\left|8(y-1)^2-(y-1)^2ight|=3 \lambda^2 \Rightarrow(y-1)^2=\left(\frac{\sqrt{3} \lambda}{\sqrt{7}}ight)^2$

$y-1= \pm \frac{\sqrt{3} \lambda}{\sqrt{7}} \Rightarrow b_1=1+\frac{\sqrt{3 \lambda}}{\sqrt{7}}, b_2=1-\frac{\sqrt{3} \lambda}{\sqrt{17}}$

$D =5\left(\frac{2 \sqrt{3} \lambda}{\sqrt{7}}ight)^2=\frac{5 	imes 4 	imes 3 \lambda^2}{7}=\frac{5 	imes 4 	imes 27}{7}=77.14$

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