Animated Solution for Mathematics - Conic Sections: Comprehension Passage

A Journey into Coordinate Geometry

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of coordinate geometry.

Often, when we see a problem involving loci and distances, our instinct is to panic and start calculating coordinates immediately. But I want you to pause. Take a breath. Let us look at the geometry first, and the algebra will follow like a faithful servant.

The Birth of the Locus

We begin with two lines, $L_1:x\sqrt{2}+y-1=0$ and $L_2:x\sqrt{2}-y+1=0$. These are our anchors. We are looking for a point $P(x,y)$ such that the product of its distances from these lines is a constant, ${\lambda }^2$.

Using the standard perpendicular distance formula, we find the distance $d_1$ from $L_1$ and $d_2$ from $L_2$. The denominator for both is $\sqrt{(\sqrt{2}{)}^2+(1{)}^2}=\sqrt{3}$.

When we multiply these distances, we get:

Look at that numerator! It is a classic difference of squares waiting to happen. By grouping terms, we transform this into:

This is the soul of our locus $C$. It is not just an equation; it is a curve that bends and shapes itself based on the value of $\lambda$.

The Intersection and the Slope Trick

A new line, $y=2x+1$, cuts through our locus. Instead of solving for $x$ and $y$ separately, notice that $y-1=2x$. This is a gift!

Substituting this into our locus equation, we get:

This tells us that the intersection points $R$ and $S$ are symmetric about the $y$-axis. Now, for the distance $RS$. Do not calculate the coordinates! Use the slope.

Since the line has a slope of $2$, the distance between any two points on it is $\sqrt{1+m^2}\cdot |x_1-x_2|$, which is $\sqrt{5}\cdot |x_1-x_2|$. With our $x$-coordinates being $\pm \sqrt{\frac{3{\lambda }^2}{2}}$, the difference is $2\sqrt{\frac{3{\lambda }^2}{2}}=\sqrt{6{\lambda }^2}$.

Thus, $RS=\sqrt{5}\cdot \sqrt{6{\lambda }^2}=\sqrt{30{\lambda }^2}$. Equating this to $\sqrt{270}$, we find ${\lambda }^2=9$. We have cracked the first code!

The Perpendicular Bisector

Now, we find the perpendicular bisector of $RS$. The midpoint $T$ of $RS$ is $(0,1)$. The slope of $RS$ is $2$, so the slope of the bisector is $-1/2$.

The equation of this bisector is $y-1=-1/2(x-0)$, or $x=-2(y-1)$. We substitute this back into our updated locus equation $|2x^2-(y-1{)}^2|=27$.

Substituting $x=-2(y-1)$, we get:

This gives us the $y$-coordinates of $R^{\prime }$ and $S^{\prime }$. Finally, we calculate $D$, the square of the distance between $R^{\prime }$ and $S^{\prime }$.

Using the same slope trick, $D=5(y_1^{\prime }-y_2^{\prime }{)}^2$. Since $(y-1{)}^2=27/7$, the difference $(y_1^{\prime }-y_2^{\prime }{)}^2$ is $4\cdot (27/7)=108/7$.

Thus, $D=5\cdot (108/7)=540/7\approx 77.14$. We have navigated the geometry, conquered the algebra, and arrived at the truth. Remember, the math is not just about the numbers; it is about the story they tell.