Let ABCD be a trapezium whose vertices lie on the parabola y²=4x Let the sides AD and BC of the trap

Q: Wait, why is the diagonal AC a focal chord? The problem just says it passes through (1,0).

That is a brilliant observation! In the geometry of a parabola $y^2 = 4ax$, the focus is defined as the point $(a, 0)$. For our parabola $y^2 = 4x$, $a=1$, so the focus is indeed $(1, 0)$. Any chord passing through the focus is, by definition, a focal chord. Recognizing this immediately simplifies the problem, as it unlocks the powerful focal chord properties.

Q: Why is the product of parameters $t_1 t_2 = -1$?

This is a standard property derived from the slope of the chord. If you take two points $A(at_1^2, 2at_1)$ and $C(at_2^2, 2at_2)$ and force the line connecting them to pass through $(a, 0)$, the slope condition $\frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2at_1 - 0}{at_1^2 - a}$ leads directly to $t_1 t_2 = -1$. It is a beautiful result that saves you from tedious coordinate geometry!

Q: Can we solve this without using parameters?

Technically, yes, but it would be incredibly messy. You would have to work with the equation of the line passing through $(1, 0)$ and intersect it with the parabola, leading to a quadratic equation in $x$ or $y$. Using parameters $t$ is the 'JEE way'—it transforms a complex algebraic problem into a clean, elegant geometric one. Trust the parameters!

Question

Let ABCD be a trapezium whose vertices lie on the parabola $y^{2}=4x$ Let the sides AD and BC of the trapezium be parallel to to y-axis. If the diagonal AC is of length $\frac{25}{4}$ and it passes through the point (1,0), then the area of ABCD is

Options:

(A) $\frac{75}{4}$

(B) $\frac{25}{2}$

(C) $\frac{125}{8}$

(D) $\frac{75}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are not just solving a problem; we are uncovering the hidden symmetry of a trapezium dancing on the curve of a parabola.

Imagine you are standing on the Cartesian plane, looking at the curve $y^2 = 4x$. This is the simplest, most elegant parabola, with its vertex at the origin and its focus sitting proudly at $S(1, 0)$.

The problem asks us to consider a trapezium $ABCD$ inscribed within this curve. The sides $AD$ and $BC$ are parallel to the $y$-axis, which is our first major clue. This tells us that the trapezium is symmetric about the $x$-axis. If $A$ has a coordinate $(x_A, y_A)$, then $D$ must be $(x_A, -y_A)$. This symmetry is our best friend.

### The Power of Parameters

Instead of wrestling with raw coordinates, let us embrace the beauty of parametric form. For the parabola $y^2 = 4x$, any point can be represented as $(t^2, 2t)$.

Let the parameter for point $A$ be $t_1$. Thus, $A = (t_1^2, 2t_1)$. Because $AD$ is parallel to the $y$-axis, $D$ shares the same $x$-coordinate, making $D = (t_1^2, -2t_1)$.

We apply the same logic to the other side, $BC$. Let the parameter for point $B$ be $t_2$. Then $B = (t_2^2, 2t_2)$ and $C = (t_2^2, -2t_2)$. We have now mapped the entire trapezium using just two variables, $t_1$ and $t_2$.

### The Focal Chord Revelation

Now, look at the diagonal $AC$. The problem whispers a secret: it passes through $(1, 0)$. As we discussed, $(1, 0)$ is the focus of our parabola. This transforms $AC$ into a focal chord.

The property of a focal chord is one of the most elegant tools in your arsenal: for a chord connecting points with parameters $t_1$ and $t_2$, the product of the parameters is $t_1 t_2 = -1$. In our specific setup, since $C$ is $(t_2^2, -2t_2)$, the parameter is effectively $-t_2$.

Thus, the condition becomes $t_1 \cdot (-t_2) = -1$, which simplifies to $t_1 t_2 = 1$. This is the key that unlocks the entire problem.

### The Calculation

We are given that the length of $AC$ is $\frac{25}{4}$. The length of a focal chord with parameters $t_1$ and $t_2$ is given by the formula $a(t_1 - t_2)^2$.

Using the standard form $a(t + \frac{1}{t})^2$ with $a=1$, the length is:

$$ (t_1 + \frac{1}{t_1})^2 = \frac{25}{4} $$

Taking the square root, we find $t_1 + \frac{1}{t_1} = \frac{5}{2}$. Solving this quadratic:

$$ 2t_1^2 - 5t_1 + 2 = 0 $$

We find $t_1 = 2$ or $t_1 = \frac{1}{2}$. Let us choose $t_1 = 2$, which implies $t_2 = \frac{1}{2}$.

### The Final Area

With our parameters in hand, the coordinates fall into place like pieces of a puzzle. $A = (4, 4)$, $D = (4, -4)$, $B = (\frac{1}{4}, 1)$, and $C = (\frac{1}{4}, -1)$.

The parallel sides $AD$ and $BC$ have lengths $8$ and $2$, respectively. The height of the trapezium is the distance between the $x$-coordinates:

$$ 4 - \frac{1}{4} = \frac{15}{4} $$

The area is calculated as:

$$ 	ext{Area} = \frac{1}{2} 	imes (8 + 2) 	imes \frac{15}{4} = 5 	imes \frac{15}{4} = \frac{75}{4} $$

We have arrived at the destination. The beauty of this problem lies not in the final number, but in how the geometry of the parabola guided us to the solution. The final answer is **$\frac{75}{4}$**.

Select Answer:

Visualized Solution

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

Symmetry and Vertices $A$ and $D$

Symmetry and Vertices $B$ and $C$

The Focal Chord Property

Length of the Focal Chord

Solving for Parameter $t_{1}$

Coordinates of the Vertices

Lengths of Parallel Sides

Height of the Trapezium

Final Area Calculation

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

Analyzing the Setup

The Power of Parameters

The Focal Chord Revelation

The Calculation

The Final Area

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