If |4a² 4a 1; 4b² 4b 1; 4c² 4c 1| |f(-1); f(1); f(2)| = |3a² + 3a; 3b² + 3b; 3c² + 3c|, f(x) is a qu

Q: Wait, why can we just replace $a, b, c$ with $x$? Isn't that a huge leap?

That is a brilliant question! Think of it this way: the matrix equation gives us three distinct points where two quadratic expressions agree. A quadratic is uniquely determined by three points. If two quadratics agree at three points, they must be the same function everywhere. It is not a leap; it is a fundamental property of polynomials!

Q: Why is the vertex at $(0,1)$? I tried to find it using $-b/2a$ but got confused.

You are overthinking the algebra! Look at the values: $f(-1) = 3/4$ and $f(1) = 3/4$. Because the function values are equal at $x = -1$ and $x = 1$, the axis of symmetry must be exactly in the middle, at $x = 0$. That is the beauty of symmetry—it saves you from tedious calculations.

Q: Why do we use the slope condition $m_{VA} \cdot m_{VB} = -1$? Is there another way?

You could use the distance formula or the dot product of vectors $\vec{VA} \cdot \vec{VB} = 0$, but the slope condition is the most direct path in coordinate geometry. It is the 'language' of perpendicularity. Always look for the path of least resistance in JEE problems!

Question

If $\left|\begin{array}{ccc}4 a^{2} & 4 a & 1 \ 4 b^{2} & 4 b & 1 \ 4 c^{2} & 4 c & 1\end{array}ight|\left|\begin{array}{c}f(-1) \ f(1) \ f(2)\end{array}ight|=\left|\begin{array}{c}3 a^{2}+3 a \ 3 b^{2}+3 b \ 3 c^{2}+3 c\end{array}ight|, f(x)$ is a quadratic function and its maximum value occurs at a point $V$. $A$ is a point of intersection of $y=f(x)$ with $x$-axis and point $B$ is such that chord $A B$ subtends a right angle at $V$. Find the area enclosed by $f(x)$ and chord $A B$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to dismantle a problem that looks like a terrifying beast of linear algebra but is actually a beautiful, elegant dance of functions. When you first see that determinant, your instinct might be to expand it. Stop! Take a breath.

In JEE Advanced, if you see a matrix equation involving variables like $a$, $b$, and $c$ that all follow the same pattern, you are likely looking at a polynomial identity in disguise. Look at the given equation:

$$
\left| \begin{array}{ccc} 4a^{2} & 4a & 1 \ 4b^{2} & 4b & 1 \ 4c^{2} & 4c & 1 \end{array} ight| \left| \begin{array}{c} f(-1) \ f(1) \ f(2) \end{array} ight| = \left| \begin{array}{c} 3a^{2} + 3a \ 3b^{2} + 3b \ 3c^{2} + 3c \end{array} ight|
$$

If you perform the matrix multiplication, you get three identical equations for $a$, $b$, and $c$. This means the quadratic expression $4x^2 f(-1) + 4x f(1) + f(2)$ and the polynomial $3x^2 + 3x$ agree at three distinct points ($x=a, b, c$).

Since a quadratic is uniquely determined by three points, they must be the same function! Thus, we have the identity:

$$
4x^2 f(-1) + 4x f(1) + f(2) \equiv 3x^2 + 3x
$$

### The Symmetry of the Parabola

Now that we have our identity, the rest is just peeling back the layers. By comparing the coefficients of $x^2$, $x$, and the constant term, we unlock the secrets of $f(x)$.

The coefficient of $x^2$ gives $4f(-1) = 3$, so $f(-1) = \frac{3}{4}$. The coefficient of $x$ gives $4f(1) = 3$, so $f(1) = \frac{3}{4}$. Finally, the constant term is zero, so $f(2) = 0$.

Let $f(x) = px^2 + qx + r$. Because $f(-1) = f(1) = \frac{3}{4}$, the parabola is perfectly symmetric about the $y$-axis, which immediately tells us $q = 0$. We are left with $f(x) = px^2 + r$.

Using our points, we solve the system: $p + r = \frac{3}{4}$ and $4p + r = 0$. Subtracting these gives $3p = -\frac{3}{4}$, so $p = -\frac{1}{4}$, and consequently $r = 1$. Our function is:

$$
f(x) = -\frac{1}{4}x^2 + 1
$$

This is a beautiful, downward-opening parabola with its vertex at $V(0, 1)$.

### The Geometry of the Chord

Now, let's step into the coordinate plane. We have our vertex $V(0, 1)$. We need to find point $A$, which is an $x$-intercept. Setting $f(x) = 0$, we get $-\frac{1}{4}x^2 + 1 = 0$, which leads to $x^2 = 4$, so $x = \pm 2$. Let's choose $A = (-2, 0)$.

Here is where the problem tests your geometric intuition. We have a chord $AB$ that subtends a right angle at $V$. This means the line segment $VA$ is perpendicular to $VB$.

The slope of $VA$ is $m_{VA} = \frac{1 - 0}{0 - (-2)} = \frac{1}{2}$. For $VB$ to be perpendicular, its slope $m_{VB}$ must be $-2$. If $B$ is $(x_1, y_1)$, then $\frac{1 - y_1}{0 - x_1} = -2$, which simplifies to $1 - y_1 = 2x_1$.

Since $B$ lies on the parabola, $y_1 = -\frac{1}{4}x_1^2 + 1$. Substituting this into our slope equation, the $1$s cancel out, and we get $\frac{1}{4}x_1^2 = 2x_1$. Since $B$ is not the vertex, $x_1 = 8$. Plugging this back, $y_1 = -15$. Point $B$ is $(8, -15)$.

### The Final Integration

We are in the home stretch. We need the area enclosed by the parabola and the chord $AB$. The chord passes through $A(-2, 0)$ and $B(8, -15)$. Its slope is $\frac{-15 - 0}{8 - (-2)} = -\frac{3}{2}$. The equation is $y = -\frac{3}{2}x - 3$.

The area is the integral of the upper curve (the parabola) minus the lower curve (the chord) from $x = -2$ to $x = 8$:

$$
Area = \int_{-2}^{8} \left[ \left(-\frac{1}{4}x^2 + 1ight) - \left(-\frac{3}{2}x - 3ight) ight] dx
$$

Simplifying the integrand, we get $\int_{-2}^{8} (-\frac{1}{4}x^2 + \frac{3}{2}x + 4) dx$. Integrating term by term, we get:

$$
\left[ -\frac{1}{12}x^3 + \frac{3}{4}x^2 + 4x ight]_{-2}^{8}
$$

Evaluating at the limits, we find the area to be $\frac{112}{3} - (-\frac{13}{3}) = \mathbf{\frac{125}{3}}$ square units. Take a moment to appreciate this. We started with a matrix, moved through symmetry, navigated coordinate geometry, and finished with calculus. This is the essence of JEE Advanced—connecting disparate dots to reveal a single, coherent truth. You've got this!

Visualized Solution

Analyzing the Matrix Equation

Comparing Coefficients

Determining $f (x)$

Identifying Vertex $V$ and Point $A$

Orthogonality at Vertex $V$

Setting up Slopes

Coordinates of Point $B$

Equation of Chord $A B$

Setting up the Area Integral

Simplifying the Integrand

Integrating the Function

Final Area Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Symmetry of the Parabola

The Geometry of the Chord

The Final Integration

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