Let A, B and C be three points on the parabola y² = 6x and let the line segment AB meet the line L t

Q: Wait, why do we use $at^2$ and $2at$ instead of $x$ and $y$? Won't that make it more complicated?

Great question! Using standard $(x, y)$ coordinates introduces square roots and makes the algebra messy. Parametric coordinates are specifically designed for parabolas to linearize the relationships. It turns quadratic equations into linear ones, which is exactly what we want to avoid getting lost in the weeds.

Q: Why does the final answer not depend on the parameters $t_1, t_2,$ and $t_3$? Does that mean the points $A, B,$ and $C$ can be anywhere?

Exactly! That is the beauty of this problem. The ratio is a geometric property of the parabola itself. No matter where you pick $A, B,$ and $C$, the ratio remains constant. It's a testament to the elegance of conic sections.

Q: I'm stuck on the chord equation. Is there a simpler way to write it?

The chord equation $y - 2at_1 = \frac{2}{t_1 + t_2}(x - at_1^2)$ is the standard form. You can also derive it by finding the slope between $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$, which is $\frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$. Once you have the slope, the point-slope form is your best friend.

Question

Let $A, B$ and $C$ be three points on the parabola $y^2 = 6x$ and let the line segment $AB$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $(\frac{AM \cdot BN}{CD})^2$ is equal to _________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The parabola is given by the equation $y^2 = 6x$. By comparing this to the standard form $y^2 = 4ax$, we identify $4a = 6$, which yields the parameter **$a = 1.5$**.

To simplify the geometry, we represent any point on the parabola using parametric coordinates $(at^2, 2at)$. Thus, we define points $A, B,$ and $C$ as:
$$A = (at_1^2, 2at_1), \quad B = (at_2^2, 2at_2), \quad C = (at_3^2, 2at_3)$$

### The Geometry of the Line $L$

The line $L$ passes through $C$ and is parallel to the $x$-axis. Since $C$ has a $y$-coordinate of $2at_3$, the equation of line $L$ is simply:
$$y = 2at_3$$

Points $M$ and $N$ are the feet of the perpendiculars from $A$ and $B$ to line $L$. The lengths of these perpendicular segments are the vertical distances from the points to the line:
$$AM = |2at_1 - 2at_3| = 2a|t_1 - t_3|$$
$$BN = |2at_2 - 2at_3| = 2a|t_2 - t_3|$$

### The Chord $AB$ and the Intersection $D$

The equation of the chord $AB$ connecting two points on the parabola is given by:
$$y - 2at_1 = \frac{2}{t_1 + t_2}(x - at_1^2)$$

Point $D$ is the intersection of this chord with the line $L$ ($y = 2at_3$). Substituting $y = 2at_3$ into the chord equation and solving for $x_D$:
$$2at_3 - 2at_1 = \frac{2}{t_1 + t_2}(x_D - at_1^2)$$
$$a(t_3 - t_1)(t_1 + t_2) = x_D - at_1^2$$
$$x_D = a(t_1t_3 + t_2t_3 - t_1^2 - t_1t_2 + t_1^2) = a(t_1t_3 + t_2t_3 - t_1t_2)$$

### The Grand Cancellation

The distance $CD$ is the difference between the $x$-coordinates of $C$ and $D$:
$$CD = |x_D - at_3^2| = a|t_1t_3 + t_2t_3 - t_1t_2 - t_3^2|$$

Factoring the expression inside the absolute value:
$$CD = a|(t_3 - t_1)(t_2 - t_3)|$$

Now, we evaluate the ratio $\frac{AM \cdot BN}{CD}$:
$$\frac{AM \cdot BN}{CD} = \frac{(2a|t_1 - t_3|) \cdot (2a|t_2 - t_3|)}{a|t_3 - t_1| \cdot |t_2 - t_3|}$$

Since $|t_1 - t_3| = |t_3 - t_1|$, these terms cancel out completely, leaving:
$$\frac{AM \cdot BN}{CD} = \frac{4a^2}{a} = 4a$$

### Final Calculation

Given $a = 1.5$, the ratio is $4(1.5) = 6$. The problem asks for the square of this ratio:
$$6^2 = \mathbf{36}$$

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola and Points

Identifying the Parabola Parameter $a$

Parametric Representation of Points

Defining the Horizontal Line $L$

Calculating Length $A M$

Calculating Length $B N$

Equation of the Chord $A B$

Finding the Intersection Point $D$

Solving for $x_{D}$

Calculating Length $C D$

Forming the Ratio

Simplifying the Expression

Final Calculation and Conclusion

The Sigma Insight: Parametric Equations

Analyzing the Setup

The Geometry of the Line $L$

The Chord $A B$ and the Intersection $D$

The Grand Cancellation

Final Calculation

Similar Questions

A line passing through the point $P (5, 5)$ intersects the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1$ at $A$ and $B$ such that $(P A) \cdot (P B)$ is maximum. Then $5 (P A^{2} + P B^{2})$ is equal to :

Let $O$ be the vertex and $Q$ be any point on the parabola, $x^{2} = 8 y$ . If the point $P$ divides the line segment $O Q$ internally in the ratio $1 : 3$ , then locus of $P$ is:

Let one end of a focal chord of the parabola $y^{2} = 16 x$ be $(16, 16)$ . If $P (α, β)$ divides this focal chord internally in the ratio $5 : 2$ , then the minimum value of $α + β$ is equal to :

Enter Numerical Value:

Visualized Solution

Visualizing the Parabola and Points

Identifying the Parabola Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Parametric Representation of Points

Defining the Horizontal Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Calculating Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM

Calculating Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BN

Equation of the Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xD​

Calculating Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CD