Consider the lines L_1 and L_2 defined by L_1 | vedclass

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(A) Let $P(x, y)$. The distances from $L_1$ and $L_2$ are $d_1 = \frac{|x\sqrt{2} + y - 1|}{\sqrt{2+1}} = \frac{|x\sqrt{2} + y - 1|}{\sqrt{3}}$ and $d

Vedclass · Accepted Answer

$9, 77.14$

Vedclass · Answer

(A) Let $P(x, y)$. The distances from $L_1$ and $L_2$ are $d_1 = \frac{|x\sqrt{2} + y - 1|}{\sqrt{2+1}} = \frac{|x\sqrt{2} + y - 1|}{\sqrt{3}}$ and $d_2 = \frac{|x\sqrt{2} - y + 1|}{\sqrt{2+1}} = \frac{|x\sqrt{2} - y + 1|}{\sqrt{3}}$.Given $d_1 d_2 = \lambda^2$,so $\frac{|(x\sqrt{2})^2 - (y-1)^2|}{3} = \lambda^2$,which gives $|2x^2 - (y-1)^2| = 3\lambda^2$.For the line $y = 2x + 1$,substitute $y-1 = 2x$ into the locus equation: $|2x^2 - (2x)^2| = 3\lambda^2 \Rightarrow |-2x^2| = 3\lambda^2 \Rightarrow x^2 = \frac{3\lambda^2}{2}$.The points $R$ and $S$ have $x$-coordinates $x_1 = \sqrt{\frac{3\lambda^2}{2}}$ and $x_2 = -\sqrt{\frac{3\lambda^2}{2}}$.The distance $RS = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} = \sqrt{(x_1-x_2)^2 + (2(x_1-x_2))^2} = \sqrt{5}|x_1-x_2| = \sqrt{5} \cdot 2\sqrt{\frac{3\lambda^2}{2}} = \sqrt{5} \cdot \sqrt{6\lambda^2} = \sqrt{30\lambda^2}$.Given $\sqrt{30\lambda^2} = \sqrt{270} \Rightarrow 30\lambda^2 = 270 \Rightarrow \lambda^2 = 9$.The midpoint $T$ of $RS$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (0, 1)$.The slope of $RS$ is $2$,so the slope of the perpendicular bisector is $-\frac{1}{2}$.The equation of the perpendicular bisector is $y - 1 = -\frac{1}{2}(x - 0) \Rightarrow x + 2y = 2 \Rightarrow x = 2 - 2y$.Substitute $x = 2(1-y)$ into $|2x^2 - (y-1)^2| = 3\lambda^2 = 27$:$|2(4(1-y)^2) - (y-1)^2| = 27 \Rightarrow |8(y-1)^2 - (y-1)^2| = 27 \Rightarrow 7(y-1)^2 = 27 \Rightarrow (y-1)^2 = \frac{27}{7}$.The distance $D = (x_1^{\prime}-x_2^{\prime})^2 + (y_1^{\prime}-y_2^{\prime})^2 = 5(y_1^{\prime}-y_2^{\prime})^2 = 5(4(y-1)^2) = 20 \cdot \frac{27}{7} = \frac{540}{7} \approx 77.14$.

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