Let A be the focus of the parabola y² = 8x. Let the line y = mx + c intersect the parabola at two di

Q: Why do we use parametric coordinates? Isn't it easier to just solve the intersection of the line and the parabola?

That is a classic trap! If you substitute $y = mx + c$ into $y^2 = 8x$, you get a quadratic in $x$ with coefficients involving $m$ and $c$. You would then have to deal with the centroid formula in terms of $m$ and $c$, which is a nightmare. Parametric coordinates allow us to work with the parameters $t_1$ and $t_2$ directly, which are much more algebraically friendly and symmetric.

Q: Wait, do I need to find the values of $t_1$ and $t_2$ separately?

Absolutely not! In fact, trying to solve for $t_1$ and $t_2$ individually is a waste of time. Notice how the centroid formula and the distance formula only require the sum and the difference (or product) of the parameters. Always look for symmetric expressions—they are the secret language of coordinate geometry.

Q: Is there a specific reason we calculate $(t_1 - t_2)^2$ instead of just $(t_1 - t_2)$?

Excellent observation! The distance formula involves squares of differences. By calculating $(t_1 - t_2)^2$ directly, we avoid dealing with square roots and potential sign errors. It keeps the math clean, elegant, and robust.

Question

Let $A$ be the focus of the parabola $y^2 = 8x$. Let the line $y = mx + c$ intersect the parabola at two distinct points $B$ and $C$. If the centroid of the triangle $ABC$ is $\left(\frac{7}{3}, \frac{4}{3}\right)$, then $(BC)^2$ is equal to :

Options:

(A) 80

(B) 41

(C) 89

(D) 32

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of the parabola $y^2 = 8x$ provides the foundation for our problem. By comparing this to the standard form $y^2 = 4ax$, we identify $4a = 8$, which yields $a = 2$.

The focus of a parabola $y^2 = 4ax$ is located at $(a, 0)$. Therefore, our anchor point, the focus $A$, is at $(2, 0)$.

### The Power of Parameters

To avoid the algebraic clutter of linear equations, we utilize parametric coordinates. For a parabola $y^2 = 4ax$, any point can be represented as $(at^2, 2at)$.

With $a = 2$, our points $B$ and $C$ on the parabola are defined as:
$$B = (2t_1^2, 4t_1)$$
$$C = (2t_2^2, 4t_2)$$

By adopting this parametric approach, we reduce the problem to managing the variables $t_1$ and $t_2$ rather than solving for complex coordinate intersections.

### The Centroid Bridge

We are given the centroid $G = (\frac{7}{3}, \frac{4}{3})$. The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:
$$G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right)$$

Applying this to triangle $ABC$ using the $y$-coordinates:
$$\frac{0 + 4t_1 + 4t_2}{3} = \frac{4}{3} \Rightarrow t_1 + t_2 = 1$$

Applying this to the $x$-coordinates:
$$\frac{2 + 2t_1^2 + 2t_2^2}{3} = \frac{7}{3} \Rightarrow 2 + 2(t_1^2 + t_2^2) = 7 \Rightarrow t_1^2 + t_2^2 = \frac{5}{2}$$

### The Algebraic Symphony

We now determine the values required for the distance formula. Using the identity $(t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2$, we substitute our known values:
$$1^2 = \frac{5}{2} + 2t_1t_2 \Rightarrow 2t_1t_2 = -\frac{3}{2} \Rightarrow t_1t_2 = -\frac{3}{4}$$

Next, we calculate $(t_1 - t_2)^2$ using the identity $(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1t_2$:
$$(t_1 - t_2)^2 = (1)^2 - 4\left(-\frac{3}{4}\right) = 1 + 3 = 4$$

### Final Calculation

The square of the distance $BC$ is given by $BC^2 = (x_B - x_C)^2 + (y_B - y_C)^2$. Substituting our parametric forms:
$$BC^2 = (2t_1^2 - 2t_2^2)^2 + (4t_1 - 4t_2)^2$$
$$BC^2 = 4(t_1^2 - t_2^2)^2 + 16(t_1 - t_2)^2$$

Since $(t_1^2 - t_2^2)^2 = (t_1 - t_2)^2(t_1 + t_2)^2$, we substitute the values derived in the previous steps:
$$BC^2 = 4(t_1 - t_2)^2(t_1 + t_2)^2 + 16(t_1 - t_2)^2$$
$$BC^2 = 4(4)(1)^2 + 16(4) = 16 + 64 = 80$$

The square of the distance between points $B$ and $C$ is **80**.

Animated Solution for Mathematics - Conic Sections: Let $A$ be the focus of the parabola $y^{2} = 8 x$ . Let the line $y = m x + c$ intersect the parabola at two distinct points $B$ and $C$ . If the centroid of the triangle $A B C$ is $(\frac{7}{3}, \frac{4}{3})$ , then $(B C)^{2}$ is equal to :

Select Answer:

Visualized Solution

Finding the Focus $A$

Parametric Coordinates of $B$ and $C$

Centroid of Triangle $A B C$

Using the Centroid's $y$ -coordinate

Using the Centroid's $x$ -coordinate

Finding the Product $t_{1} t_{2}$

Finding the Difference $(t_{1} - t_{2})^{2}$

Setting up the Distance $B C^{2}$

Simplifying and Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of Parameters

The Centroid Bridge

The Algebraic Symphony

Final Calculation

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Select Answer:

Visualized Solution

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

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Centroid of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Using the Centroid's .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate

Using the Centroid's .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate

Finding the Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​t2​

Finding the Difference .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (t1​−t2​)2

Setting up the Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC2