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Step-by-Step Solution
Step 1: Identify the parabola and given points
The equation of the given parabola is $y^2 = 4x$. Its vertex is at $O(0, 0)$. We have two points on this parabola:
• $P(4, -4)$, • $Q(9, 6)$.
Step 2: Parametric form of a general point on the parabola
A standard parametric form for any point on $y^2 = 4x$ is $(t^2, 2t)$, where $t$ is a real parameter. Thus, let the point $X$ on the parabola be $X(t^2, 2t)$.
Step 3: Express the coordinates of P and Q in parametric form (for reference)
Observe that:
• For $P(4, -4)$, we can match it with $(t^2, 2t)$: $t^2 = 4$ and $2t = -4 \implies t = -2$.
• For $Q(9, 6)$, we can match it with $(t^2, 2t)$: $t^2 = 9$ and $2t = 6 \implies t = 3$.
These checks confirm that P and Q indeed lie on $y^2 = 4x$.
Step 4: Area formula for triangle PXQ
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is given by:
Area=12\bigl|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)\bigr|.
Taking $P(4, -4)$, $Q(9, 6)$, and $X(t^2, 2t)$, we substitute into this determinant formula.
Step 5: Substitute coordinates into the area determinant
Let $X = (t^2, 2t)$. Then the area of $\triangle PXQ$ is:
Area=12\bigl|4(6-2t)+9(2t+4)+t2(-4-6)\bigr|.
Simplify step by step:
$ 4(6 - 2t) = 24 - 8t, $
$ 9(2t + 4) = 18t + 36, $
$ t^2(-4 - 6) = -10t^2. $
So inside the absolute value we get:
$ 24 - 8t + 18t + 36 - 10t^2 = 60 + 10t - 10t^2. $
Hence, $ \text{Area} = \frac{1}{2} \bigl| 60 + 10t - 10t^2 \bigr| = \frac{1}{2} \cdot 10 \bigl| 6 + t - t^2 \bigr| = 5 \bigl| 6 + t - t^2 \bigr|. $
Step 6: Maximize the expression for area
We focus on maximizing the expression $6 + t - t^2$. Because area uses an absolute value, we look for where $t$ lies on the arc joining $P$ and $Q$ (i.e., $t \in [-2, 3]$).
• The quadratic $-t^2 + t + 6$ opens downward (since the coefficient of $t^2$ is negative).
• Its vertex occurs at $t = -\frac{b}{2a} = -\frac{1}{2 \cdot (-1)} = \frac{1}{2}.$
This $t = \frac{1}{2}$ is within the interval $[-2, 3]$, so it gives the maximum of $6 + t - t^2$.
Step 7: Find the maximum value of the quadratic part
At $t = \frac{1}{2}$,
$ 6 + \frac{1}{2} - \left(\frac{1}{2}\right)^2 = 6 + 0.5 - 0.25 = 6.25 = \frac{25}{4}. $
Therefore, the maximum expression inside the absolute value is $ \frac{25}{4}. $
Hence, the maximum area becomes:
$ \text{Area}_{\max} = 5 \times \frac{25}{4} = \frac{125}{4}. $
Step 8: Final answer
The maximum area of $\triangle PXQ$ is $ \frac{125}{4} \text{ (sq. units)}. $
Step-by-Step Solution
Step 1: Identify the parabola and given points
The equation of the given parabola is $y^2 = 4x$. Its vertex is at $O(0, 0)$. We have two points on this parabola:
• $P(4, -4)$, • $Q(9, 6)$.
Step 2: Parametric form of a general point on the parabola
A standard parametric form for any point on $y^2 = 4x$ is $(t^2, 2t)$, where $t$ is a real parameter. Thus, let the point $X$ on the parabola be $X(t^2, 2t)$.
Step 3: Express the coordinates of P and Q in parametric form (for reference)
Observe that:
• For $P(4, -4)$, we can match it with $(t^2, 2t)$: $t^2 = 4$ and $2t = -4 \implies t = -2$.
• For $Q(9, 6)$, we can match it with $(t^2, 2t)$: $t^2 = 9$ and $2t = 6 \implies t = 3$.
These checks confirm that P and Q indeed lie on $y^2 = 4x$.
Step 4: Area formula for triangle PXQ
The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is given by:
Area=12\bigl|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)\bigr|. \text{Area} = \frac{1}{2} \bigl| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigr|.
Taking $P(4, -4)$, $Q(9, 6)$, and $X(t^2, 2t)$, we substitute into this determinant formula.
Step 5: Substitute coordinates into the area determinant
Let $X = (t^2, 2t)$. Then the area of $\triangle PXQ$ is:
Area=12\bigl|4(6-2t)+9(2t+4)+t2(-4-6)\bigr|. \text{Area} = \frac{1}{2} \bigl| 4(6 - 2t) + 9(2t + 4) + t^2(-4 - 6) \bigr|.
Simplify step by step:
$ 4(6 - 2t) = 24 - 8t, $
$ 9(2t + 4) = 18t + 36, $
$ t^2(-4 - 6) = -10t^2. $
So inside the absolute value we get:
$ 24 - 8t + 18t + 36 - 10t^2 = 60 + 10t - 10t^2. $
Hence, $ \text{Area} = \frac{1}{2} \bigl| 60 + 10t - 10t^2 \bigr| = \frac{1}{2} \cdot 10 \bigl| 6 + t - t^2 \bigr| = 5 \bigl| 6 + t - t^2 \bigr|. $
Step 6: Maximize the expression for area
We focus on maximizing the expression $6 + t - t^2$. Because area uses an absolute value, we look for where $t$ lies on the arc joining $P$ and $Q$ (i.e., $t \in [-2, 3]$).
• The quadratic $-t^2 + t + 6$ opens downward (since the coefficient of $t^2$ is negative).
• Its vertex occurs at $t = -\frac{b}{2a} = -\frac{1}{2 \cdot (-1)} = \frac{1}{2}.$
This $t = \frac{1}{2}$ is within the interval $[-2, 3]$, so it gives the maximum of $6 + t - t^2$.
Step 7: Find the maximum value of the quadratic part
At $t = \frac{1}{2}$,
$ 6 + \frac{1}{2} - \left(\frac{1}{2}\right)^2 = 6 + 0.5 - 0.25 = 6.25 = \frac{25}{4}. $
Therefore, the maximum expression inside the absolute value is $ \frac{25}{4}. $
Hence, the maximum area becomes:
$ \text{Area}_{\max} = 5 \times \frac{25}{4} = \frac{125}{4}. $
Step 8: Final answer
The maximum area of $\triangle PXQ$ is $ \frac{125}{4} \text{ (sq. units)}. $
