Points A, B and C lie on the parabola y² = 4ax. The tangents to the parabola at A, B and C, taken in

Q: Wait, why do we use $t$ instead of just $x$ and $y$?

Using $x$ and $y$ leads to messy square roots and complex algebraic manipulation. Parameters turn geometry into elegant algebra. By using $t$, we reduce the parabola's complexity to simple polynomial expressions, which makes the area calculation much cleaner.

Q: Is the area formula for $\triangle PQR$ different from $\triangle ABC$?

Not at all! The area formula for a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is universal: $\Delta = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. The only thing that changes is the input coordinates. Once you plug in the intersection points, the math will naturally follow the same structure.

Q: Does the ratio change if I pick different points on the parabola?

That is the beauty of this problem! The ratio is a constant property of the parabola. No matter where you place points $A, B,$ and $C$, the algebraic terms involving the parameters will always cancel out in the final ratio, leaving you with the elegant result of $2:1$.

Question

Points $A, B$ and $C$ lie on the parabola $y^2 = 4ax$. The tangents to the parabola at $A, B$ and $C$, taken in pairs, intersect at points $P, Q$ and $R$. Determine the ratio of the areas of the triangles $ABC$ and $PQR$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering a fundamental truth about the parabola. When you look at the equation $y^2 = 4ax$, do not just see a curve. See a playground of symmetry and hidden ratios.

We are going to explore the relationship between three points on this curve and the triangle formed by their tangents.

### The Power of Parameters

Our journey begins with the points $A, B,$ and $C$. If we try to work with standard Cartesian coordinates $(x, y)$, we will quickly find ourselves drowning in square roots. Instead, we embrace the parametric form.

Any point on our parabola can be defined as $(at^2, 2at)$. By assigning parameters $t_1, t_2,$ and $t_3$ to our points $A, B,$ and $C$, we transform the geometry into a beautiful algebraic dance.

### The First Triangle: $	riangle ABC$

Now, let us connect these points to form $	riangle ABC$. To find its area, we use the classic determinant formula:

$$\Delta ABC = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$

Substituting our parametric coordinates, we get:

$$\Delta ABC = \frac{1}{2} |at_1^2(2at_2 - 2at_3) + at_2^2(2at_3 - 2at_1) + at_3^2(2at_1 - 2at_2)|$$

As we factor out $2a^2$, the expression simplifies to $a^2 |t_1^2(t_2 - t_3) + t_2^2(t_3 - t_1) + t_3^2(t_1 - t_2)|$. This cyclic expression is a classic in JEE mathematics, factorizing perfectly into:

$$\Delta ABC = a^2 |(t_1-t_2)(t_2-t_3)(t_3-t_1)|$$

Keep this result close; it is the foundation of our proof.

### The Tangent Dance

Now, we draw the tangents at $A, B,$ and $C$. These lines extend and intersect at points $P, Q,$ and $R$. Here is where the magic happens.

A standard property of the parabola $y^2 = 4ax$ tells us that the intersection of tangents at $t_1$ and $t_2$ is $(at_1t_2, a(t_1+t_2))$. Using this, we define:

$$P = (at_1t_2, a(t_1+t_2)), \quad Q = (at_2t_3, a(t_2+t_3)), \quad R = (at_3t_1, a(t_3+t_1))$$

### The Second Triangle: $	riangle PQR$

We now construct $	riangle PQR$ using these intersection points. Applying the same area formula, we substitute these new coordinates.

When we compute the area $\Delta PQR = \frac{1}{2} |x_P(y_Q-y_R) + x_Q(y_R-y_P) + x_R(y_P-y_Q)|$, the $a$ terms combine to form $a^2$. The differences in $y$ coordinates simplify to $a(t_2-t_1)$, $a(t_3-t_2)$, and $a(t_1-t_3)$.

Factoring out $a^2$ and the $\frac{1}{2}$ from the formula, we find:

$$\Delta PQR = \frac{a^2}{2} |t_1t_2(t_2-t_1) + t_2t_3(t_3-t_2) + t_3t_1(t_1-t_3)|$$

Just like before, this cyclic expression factorizes into the same product of differences. Thus:

$$\Delta PQR = \frac{a^2}{2} |(t_1-t_2)(t_2-t_3)(t_3-t_1)|$$

### The Grand Finale

We have arrived at the moment of truth. We want the ratio of the area of $	riangle ABC$ to the area of $	riangle PQR$.

When we divide $\Delta ABC$ by $\Delta PQR$, the entire complex product of differences cancels out completely. We are left with:

$$\frac{\Delta ABC}{\Delta PQR} = \frac{a^2 |(t_1-t_2)(t_2-t_3)(t_3-t_1)|}{\frac{a^2}{2} |(t_1-t_2)(t_2-t_3)(t_3-t_1)|} = 2$$

The ratio is **$2:1$**. This is the elegance of coordinate geometry. You have mastered the geometry; now, carry this confidence into your next challenge.

Animated Solution for Mathematics - Conic Sections: Points $A, B$ and $C$ lie on the parabola $y^{2} = 4 a x$ . The tangents to the parabola at $A, B$ and $C$ , taken in pairs, intersect at points $P, Q$ and $R$ . Determine the ratio of the areas of the triangles $A B C$ and $P QR$ .

Visualized Solution

Visualizing the Parabola and Points

Parametric Coordinates of Points

Forming $△ A B C$

Area Formula for a Triangle

Area of $△ A B C$ (Substitution)

Area of $△ A B C$ (Simplification)

Final Area of $△ A B C$

Drawing the Tangents

Intersection Points $P, Q, R$

Coordinates of Intersection Points

Forming $△ P QR$

Area of $△ P QR$ (Substitution)

Area of $△ P QR$ (Simplification)

Final Area of $△ P QR$

The Ratio of Areas

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of Parameters

The First Triangle: $△ A B C$

The Tangent Dance

The Second Triangle: $△ P QR$

The Grand Finale

Similar Questions

Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

The tangents at the point $A (1, 3)$ and $B (1, - 1)$ on the parabola $y^{2} - 2 x - 2 y = 1$ meet at the point $P$ . Then the area (in unit $^{2}$ ) of the triangle $P A B$ is :-

If the tangent at a point P on the parabola $y^{2} = 3 x$ is parallel to the line $x + 2 y = 1$ and the tangents at the points Q and R on the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1$ are perpendicular to the line $x - y = 2$ , then the area of the triangle PQR is:

$A$ is a point on the parabola $y^{2} = 4 a x$ . The normal at $A$ cuts the parabola again at point $B$ . If $A B$ subtends a right angle at the vertex of the parabola, find the slope of $A B$ .

Visualized Solution

Visualizing the Parabola and Points

Parametric Coordinates of Points

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

Area Formula for a Triangle

Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC (Substitution)

Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △ABC (Simplification)

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