A line L : y = mx + 3 meets y-axis at E(0, 3) and the arc of the parabola y² = 16x, 0 ≤ y ≤ 6 at the

Q: Why is the base of the triangle $|3 - y_1|$? I thought it was just $3 - y_1$.

Great observation! In coordinate geometry, distance is always non-negative. Since we don't know if $3 > y_1$ or $y_1 > 3$ initially, we use the absolute value $|3 - y_1|$ to ensure the area remains positive. As we solve, you'll see that for the relevant range of $y_0$, the expression simplifies naturally.

Q: Why can't we use parametric coordinates like $(4t^2, 8t)$?

You absolutely can! Parametric coordinates are a powerful tool in your arsenal. If you set $y_0 = 8t$, you'll find $x_0 = 4t^2$. The math will lead you to the exact same result. However, using $y_0$ directly is often more intuitive when the problem is already framed in terms of $y$-coordinates. Choose the path that feels most natural to you!

Q: Why do we reject $y_0 = 0$ when we solve the derivative?

Think about the physical reality of the triangle. If $y_0 = 0$, then point $F$ is at the origin $(0,0)$. If $F$ is at the origin, the triangle $EFG$ collapses into a line segment on the $y$-axis. A line segment has no area! We are looking for a triangle, so we must have a non-zero area.

Question

A line $L : y = mx + 3$ meets $y$-axis at $E(0, 3)$ and the arc of the parabola $y^2 = 16x, 0 \le y \le 6$ at the point $F(x_0, y_0)$. The tangent to the parabola at $F(x_0, y_0)$ intersects the $y$-axis at $G(0, y_1)$. The slope $m$ of the line $L$ is chosen such that the area of the triangle $EFG$ has a local maximum. Match List I with List II and select the correct answer using the code given below the lists :

List-I:

P. $m =$

Q. Maximum area of $\Delta EFG$ is

R. $y_0 =$

S. $y_1 =$

List-II:

1. 1/2

2. 4

3. 2

4. 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at the elegant curve of a parabola defined by $y^2 = 16x$. We are focusing on the upper arc, where $y$ ranges from $0$ to $6$.

Anchored on the $y$-axis is a fixed point $E(0, 3)$. A line $L$ passes through $E$ and intersects the parabola at a point $F(x_0, y_0)$. This point $F$ serves as the dynamic anchor for our geometric construction.

### The Tangent's Secret

To find the point $G(0, y_1)$ where the tangent at $F$ intersects the $y$-axis, we use the standard equation of a tangent to the parabola $y^2 = 4ax$ at $(x_0, y_0)$, which is $yy_0 = 2a(x + x_0)$. Given $y^2 = 16x$, we identify $4a = 16$, so $a = 4$.

The tangent equation becomes $yy_0 = 8(x + x_0)$. By setting $x=0$ to find the $y$-intercept $G$, we obtain:

$$y_1 y_0 = 8x_0$$

Because $F$ lies on the parabola, we know $x_0 = \frac{y_0^2}{16}$. Substituting this into our intercept equation yields the elegant relationship:

$$y_1 = \frac{y_0}{2}$$

### Constructing the Area

We now consider $\Delta EFG$. Its base lies on the $y$-axis, stretching from $E(0, 3)$ to $G(0, \frac{y_0}{2})$. The length of this base is $|3 - \frac{y_0}{2}|$.

The height of this triangle is the horizontal distance from the $y$-axis to $F$, which is simply $x_0 = \frac{y_0^2}{16}$. The area $A$ is defined by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Substituting our expressions, we get:

$$A(y_0) = \frac{1}{2} \left( 3 - \frac{y_0}{2} ight) \left( \frac{y_0^2}{16} ight)$$

This simplifies to the following function of $y_0$:

$$A(y_0) = \frac{1}{64} (6y_0^2 - y_0^3)$$

### The Calculus of Optimization

To find the local maximum, we differentiate the area function with respect to $y_0$:

$$\frac{dA}{dy_0} = \frac{1}{64} (12y_0 - 3y_0^2)$$

Setting the derivative to zero, we solve $3y_0(4 - y_0) = 0$. The critical points are $y_0 = 0$ and $y_0 = 4$.

We reject $y_0 = 0$ because it results in a degenerate triangle with no area. Thus, $y_0 = 4$ is our optimal coordinate.

### Final Calculation

With $y_0 = 4$, we find $x_0 = 1$ and $y_1 = 2$. Substituting these values back into our area formula:

$$A = \frac{1}{64} (6(4)^2 - (4)^3) = \frac{1}{64} (96 - 64) = \frac{32}{64} = \frac{1}{2}$$

The **maximum area** of the triangle $EFG$ is **$\frac{1}{2}$**.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the Parabola and Point $E$

Introducing Point $F$ and Line $L$

The Tangent and Point $G$

Equation of the Tangent

Finding the $y$ -coordinate of $G$

Expressing $y_{1}$ in terms of $y_{0}$

Visualizing Triangle $E F G$

Formulating the Area Function

Maximizing the Area

Finding the Critical Points

Calculating $x_{0}$ and Slope $m$

Calculating $y_{1}$ and Maximum Area

Final Matching

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Tangent's Secret

Constructing the Area

The Calculus of Optimization

Final Calculation

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List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the Parabola and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Introducing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

The Tangent and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G