Consider the parabola y² = 8x . Let Δ₁ be the area of the triangle formed by the end points of its l

Q: Wait, why do we reject the negative root here? Or is it just a matter of symmetry?

Great observation! In this specific problem, the parabola is symmetric about the x-axis. While the parabola itself has both positive and negative y-values, the point $P$ is given as $(\frac{1}{2}, 2)$, which is in the upper half. The latus rectum endpoints $L$ and $L'$ are at $(2, 4)$ and $(2, -4)$. Because the parabola is symmetric, the area calculation for the triangle $PL L'$ remains consistent regardless of whether you pick $P$ on the top or bottom, but we stick to the given coordinates to be precise.

Q: Why can't we just use energy conservation or some other physics principle?

Ah, the physicist's instinct! While energy conservation is the bedrock of mechanics, here we are firmly in the realm of pure coordinate geometry. This is a problem of pure mathematical relationships between curves and their tangents. There is no 'motion' or 'potential energy' here, just the static, elegant architecture of conic sections. Trust the geometry!

Q: Is there a faster way to find the area of the triangle formed by tangents?

Yes! There is a beautiful property in coordinate geometry: if you have three points on a parabola, the area of the triangle formed by the points is always exactly twice the area of the triangle formed by the tangents at those points. While it is excellent to derive it step-by-step as we did, knowing this property can be a powerful tool for verification in competitive exams.

Question

Consider the parabola $y^2 = 8x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P(\frac{1}{2}, 2)$ on the parabola and $\Delta_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latus rectum. Then $\frac{\Delta_1}{\Delta_2}$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the coordinate plane! Today, we are going to dissect a classic problem that bridges the gap between simple algebraic manipulation and the profound beauty of conic sections.

We are working with the parabola $y^2 = 8x$. By comparing this to the standard form $y^2 = 4ax$, we identify $4a = 8$, which gives us $a = 2$.

This tells us the focus lies at $(2, 0)$. Visualize this: a smooth, rightward-opening curve starting at the origin, with its focus sitting comfortably at $x = 2$. This is our playground.

### The Latus Rectum and the Point $P$

Next, we locate the latus rectum. It is the vertical chord passing through the focus, perpendicular to the axis of symmetry.

With $a = 2$, the endpoints $L$ and $L'$ are at $(a, 2a)$ and $(a, -2a)$, which translates to $L(2, 4)$ and $L'(2, -4)$.

Now, consider the point $P(\frac{1}{2}, 2)$. Checking the curve equation:
$$y^2 = 2^2 = 4$$
$$8x = 8\left(\frac{1}{2}ight) = 4$$

The equality holds! $P$ is a point on the upper half of the parabola, nestled between the vertex and the endpoint $L$.

### Calculating $\Delta_1$: The Geometric Approach

Now, we form our first triangle, $\Delta_1$, by connecting $P$, $L$, and $L'$. To find its area, we use the formula:
$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

Let the vertical segment $LL'$ be our base. Its length is $4 - (-4) = 8$ units.

The height is the horizontal distance from $P$ to the line $x = 2$. This is $|2 - \frac{1}{2}| = \frac{3}{2}$ units.

Thus, the area is:
$$\Delta_1 = \frac{1}{2} 	imes 8 	imes \frac{3}{2} = 6 	ext{ square units}$$

### The Tangent Web: An Algebraic Dance

Now, the problem shifts. We must draw tangents at $P$, $L$, and $L'$. The equation of a tangent to $y^2 = 8x$ at $(x_1, y_1)$ is $yy_1 = 4(x + x_1)$.

Applying this:
1. At $P(\frac{1}{2}, 2)$: $2y = 4(x + \frac{1}{2}) \Rightarrow y = 2x + 1$
2. At $L(2, 4)$: $4y = 4(x + 2) \Rightarrow y = x + 2$
3. At $L'(2, -4)$: $-4y = 4(x + 2) \Rightarrow y = -x - 2$

Solving these equations pairwise reveals the vertices of the second triangle, $\Delta_2$: $A(1, 3)$, $B(-1, -1)$, and $C(-2, 0)$.

### The Grand Reveal

With the vertices of $\Delta_2$ in hand, we use the coordinate area formula:
$$\Delta_2 = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substituting our coordinates $(1, 3)$, $(-1, -1)$, and $(-2, 0)$:
$$\Delta_2 = \frac{1}{2} |1(-1 - 0) + (-1)(0 - 3) + (-2)(3 - (-1))|$$
$$\Delta_2 = \frac{1}{2} |-1 + 3 - 8| = \frac{1}{2} |-6| = 3$$

Finally, the ratio is:
$$\frac{\Delta_1}{\Delta_2} = \frac{6}{3} = 2$$

This is the magic of the parabola: the triangle formed by the points is always twice the area of the triangle formed by their tangents. You have just witnessed a fundamental truth of geometry.

Enter Numerical Value:

Visualized Solution

Identify the Parabola and its Focus

Endpoints of the Latus Rectum

Locate Point $P$ on the Parabola

Calculate Area $Δ_{1}$

Equations of Tangents

Intersection of Tangents

Calculate Area $Δ_{2}$

Final Ratio and Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Latus Rectum and the Point $P$

Calculating $Δ_{1}$

The Tangent Web

The Grand Reveal

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