Let A(4,-4) and B(9,6) be points on the parabola, y² = 4x. Let C be chosen on the arc AOB of the par

Q: Why can't we just use the perpendicular distance formula from a point to a line?

You absolutely can! However, using the perpendicular distance formula requires finding the equation of the line AB, calculating the distance from a general point $(t^2, 2t)$ to that line, and then maximizing that distance. While mathematically sound, it often leads to more tedious algebraic manipulation. Using the parametric coordinate approach with the determinant area formula is generally more elegant and less prone to calculation errors in competitive exams.

Q: How do I know if the tangent is parallel to the chord?

This is a beautiful geometric property of parabolas. For any chord connecting two points on a parabola, there is exactly one point on the arc between them where the tangent is parallel to that chord. This point is the vertex of the triangle with the maximum area. You can verify this by calculating the slope of the chord AB and equating it to the derivative of the parabola $y = 2\sqrt{x}$ or $x = y^2/4$ at the point C.

Q: Why did we drop the absolute value sign when defining the area function?

Great observation! We dropped the absolute value because we analyzed the quadratic expression inside it, which was $-t^2 + t + 6$. We found its roots to be $t = -2$ and $t = 3$. Since our point C is restricted to the arc AOB, our parameter $t$ must lie strictly within the interval $(-2, 3)$. Within this interval, the downward-opening parabola $-t^2 + t + 6$ is always positive, allowing us to remove the absolute value bars without changing the function's value.

Question

Let $A(4,-4)$ and $B(9,6)$ be points on the parabola, $y^2 = 4x$. Let $C$ be chosen on the arc $AOB$ of the parabola, where $O$ is the origin, such that the area of $\Delta ACB$ is maximum. Then, the area (in sq. units) of $\Delta ACB$, is:

Options:

(A) $31 \frac{3}{4}$

(B) 32

(C) $30 \frac{1}{2}$

(D) $31 \frac{1}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the elegant curve of the parabola $y^2 = 4x$. You have two fixed anchors, point $A(4, -4)$ and point $B(9, 6)$, connected by a straight chord.

Your mission is to find a point $C$ on the arc $AOB$ that pushes the area of $\Delta ACB$ to its absolute limit. This is not just a calculation; it is a dance between geometry and calculus.

### The Power of Parametric Coordinates

To conquer this problem, we must first speak the language of the parabola. Instead of dealing with messy $x$ and $y$ variables, we use the parametric form.

For any parabola $y^2 = 4ax$, we can describe any point as $(at^2, 2at)$. Here, with $a=1$, our point $C$ becomes $(t^2, 2t)$.

This transformation is our secret weapon. It reduces the complexity of the curve to a single variable, $t$. By identifying the parameters for $A$ and $B$—where $A$ corresponds to $t = -2$ and $B$ to $t = 3$—we define the playground for our point $C$: the open interval $t \in (-2, 3)$.

### Constructing the Area Function

Now, we invoke the classic determinant formula for the area of a triangle:

$$Area = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$

Substituting our coordinates $A(4, -4)$, $B(9, 6)$, and $C(t^2, 2t)$ into this formula, we embark on an algebraic journey. As we expand the terms, we find ourselves with the expression:

$$Area = \frac{1}{2} |4(6 - 2t) + 9(2t - (-4)) + t^2(-4 - 6)|$$

Simplifying this, we arrive at the quadratic function:

$$S(t) = 5 |-t^2 + t + 6|$$

This is the heartbeat of our problem.

### The Calculus of Optimization

We have a function, and we want to find its peak. We know that within our interval $t \in (-2, 3)$, the expression $-t^2 + t + 6$ is always positive, so we can safely discard the absolute value.

To find the maximum, we take the derivative:

$$S'(t) = 5(1 - 2t)$$

Setting this to zero yields $t = \frac{1}{2}$. This is the moment of truth.

Geometrically, this result is profound: at $t = \frac{1}{2}$, the tangent to the parabola is perfectly parallel to the chord $AB$. The slope of $AB$ is $2$, and the slope of the tangent at $C$ is also $2$.

### The Final Triumph

With $t = \frac{1}{2}$ in hand, we return to our area function $S(t) = 5(6 + t - t^2)$. Substituting $t = \frac{1}{2}$, we calculate:

$$S\left(\frac{1}{2}\right) = 5\left(6 + \frac{1}{2} - \frac{1}{4}\right)$$

Simplifying this, we get:

$$5\left(\frac{24 + 2 - 1}{4}\right) = 5\left(\frac{25}{4}\right) = \frac{125}{4}$$

Thus, the **maximum area is $31 \frac{1}{4}$ square units**. You have successfully navigated the geometry, the algebra, and the calculus to find the peak.

Animated Solution for Mathematics - Conic Sections: Let $A (4, - 4)$ and $B (9, 6)$ be points on the parabola, $y^{2} = 4 x$ . Let $C$ be chosen on the arc $A O B$ of the parabola, where $O$ is the origin, such that the area of $Δ A C B$ is maximum. Then, the area (in sq. units) of $Δ A C B$ , is:

Select Answer:

Visualized Solution

Visualizing the Parabola $y^{2} = 4 x$

Parametric Representation of Point $C$

Parameter Range for Arc $A O B$

Area Formula Setup

Substituting the Coordinates

Expanding the Terms

Simplifying the Quadratic

Analyzing the Modulus

Maximizing the Area Function

Geometric Meaning of Maximum Area

Calculating Maximum Area

Final Answer

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

Analyzing the Setup

The Power of Parametric Coordinates

Constructing the Area Function

The Calculus of Optimization

The Final Triumph

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