Let P(4, -4) and Q(9, 6) be two points on the parabola, y²=4x and let X be any point on the arc POQ

Q: Wait, why do we reject the negative root when solving for the tangent slope?

Great observation! In this specific problem, we are looking for a point $X$ on the arc $POQ$. Since the chord $PQ$ lies in the first and fourth quadrants, and the arc $POQ$ is defined by the parabola $y^2=4x$ between $x=0$ and $x=9$, we must ensure our point $X$ actually lies on that arc. When we equate the slopes, we get a unique solution for $y$ that satisfies the geometry of the arc. Always check if your calculated point lies within the specified domain!

Q: Can I use the determinant formula for the area of a triangle directly?

Absolutely! The determinant formula is a powerful tool. Once you have identified the coordinates of $P$, $Q$, and the optimal point $X$, the determinant formula is often less prone to algebraic errors than the base-height method. Just be careful with the signs of the coordinates and the absolute value at the end.

Q: Is there a shortcut using the properties of the parabola?

Yes, there is! For any parabola $y^2=4ax$, the point on the arc between two points $P$ and $Q$ that maximizes the area of $\Delta PXQ$ always has a tangent parallel to the chord $PQ$. This is a standard result in coordinate geometry. Knowing this property saves you from having to derive the condition from scratch during the exam.

Question

Let $P(4, -4)$ and $Q(9, 6)$ be two points on the parabola, $y^2=4x$ and let $X$ be any point on the arc $POQ$ of this parabola, where $O$ is the vertex of this parabola, such that the area of $\Delta PXQ$ is maximum. Then this maximum area (in sq. units) is :

Options:

(A) $125/4$

(B) $125/2$

(C) $625/4$

(D) $75/2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at the elegant curve of the parabola $y^2 = 4x$. We are given two fixed points, $P(4, -4)$ and $Q(9, 6)$, resting on this curve.

Our mission is to find a point $X$ on the arc $PQ$ such that the triangle formed by $P$, $X$, and $Q$ has the largest possible area. This is a classic optimization problem that bridges geometry and calculus.

### The Geometric Intuition

The area of any triangle is defined by the formula:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

In our case, the base is the line segment $PQ$. Since $P$ and $Q$ are fixed, the length of the base $PQ$ is constant.

To maximize the area, we simply need to maximize the perpendicular height from the variable point $X$ to the line $PQ$. The maximum distance occurs at the exact moment when the tangent to the parabola at $X$ becomes perfectly parallel to the chord $PQ$.

### The Calculus Connection

First, let us find the slope of the chord $PQ$. Using the slope formula $m = \frac{y_2 - y_1}{x_2 - x_1}$, we substitute our coordinates:

$$m = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$$

Now, we turn to the parabola $y^2 = 4x$. By differentiating implicitly with respect to $x$, we get $2y \frac{dy}{dx} = 4$, which simplifies to:

$$\frac{dy}{dx} = \frac{2}{y}$$

This expression gives us the slope of the tangent at any point $(x, y)$ on the parabola. Since we need the tangent to be parallel to the chord, we set the slope of the tangent equal to the slope of the chord:

$$\frac{2}{y} = 2 \Rightarrow y = 1$$

Substituting $y = 1$ back into the parabola equation $y^2 = 4x$, we find $1^2 = 4x$, which means $x = \frac{1}{4}$. Thus, our optimal point $X$ is $(\frac{1}{4}, 1)$.

### Final Calculation

Now that we have the coordinates of all three vertices—$P(4, -4)$, $Q(9, 6)$, and $X(\frac{1}{4}, 1)$—we calculate the maximum area using the coordinate geometry formula:

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

Substituting our values, we get:

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

Breaking this down, we have $4(5) + 9(5) + \frac{1}{4}(-10)$, which simplifies to $20 + 45 - 2.5 = 62.5$.

Finally, multiplying by $\frac{1}{2}$, we obtain the **maximum area** of **$31.25$** or **$\frac{125}{4}$ square units**.

Select Answer:

Visualized Solution

Visualizing the Parabola and Points

Forming the Triangle

The Geometric Condition for Maximum Area

Tangent Parallel to Chord

Calculating the Slope of Chord $P Q$

Finding the Tangent Slope Expression

Equating Slopes to Find $y$

Finding the $x$ -coordinate of $X$

Setting up the Area Formula

Executing the Calculation

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Geometric Intuition

The Calculus Connection

Final Calculation

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:

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$ -axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from $O (0, 0)$ to the parabola $P$ which meet $P$ at $S$ and $R$ , then the area (in sq. units) of $Δ S O R$ is equal to

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 $P QR$ .

Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to

Let $P$ be the point on the parabola $y^{2} = 4 x$ which is at the shortest distance from the center $S$ of the circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$ . Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then

Tangents are drawn to the hyperbola $4 x^{2} - y^{2} = 36$ at the points $P$ and $Q$ . If these tangents intersect at the point $T (0, 3)$ then the area (in sq. units) of $Δ P T Q$ is :

The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the x-axis is :-

Let a line $y = m x (m > 0)$ intersect the parabola, $y^{2} = x$ at a point $P$ , other than the origin. Let the tangent to it at $P$ meet the $x$ -axis at the point $Q$ . If area $(Δ O P Q) = 4$ sq. units, then $m$ is equal to _________ .

Let P be a point on the parabola, $x^{2} = 4 y$ . If the distance of P from the centre of the circle, $x^{2} + y^{2} + 6 x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at P, is :