Key Concepts and Formulas

• Locus of a Point: The set of all points satisfying a given condition.
• Distance Formula: The distance $d$ between points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Area of a Quadrilateral with Perpendicular Diagonals: If the diagonals $d_1$ and $d_2$ are perpendicular, the area is $\frac{1}{2}d_1d_2$.

Step-by-Step Solution

Step 1: Define the moving point and set up the locus equation. Let $P(x, y)$ be the moving point. We are given the points $A_1(1, 2)$ and $A_2(-2, 1)$. The sum of the squares of the distances from $P$ to $A_1$ and $A_2$ is 14. We need to express this condition mathematically. $PA_1^2 = (x-1)^2 + (y-2)^2$ $PA_2^2 = (x+2)^2 + (y-1)^2$ We are given $PA_1^2 + PA_2^2 = 14$. Substituting the distance formulas: $(x-1)^2 + (y-2)^2 + (x+2)^2 + (y-1)^2 = 14$

Step 2: Expand and simplify the equation to find the locus $f(x,y)=0$. We expand the squared terms and simplify the equation to obtain the locus of point P. $(x^2 - 2x + 1) + (y^2 - 4y + 4) + (x^2 + 4x + 4) + (y^2 - 2y + 1) = 14$ Combine like terms: $2x^2 + 2x + 2y^2 - 6y + 10 = 14$ $2x^2 + 2y^2 + 2x - 6y - 4 = 0$ Divide by 2: $x^2 + y^2 + x - 3y - 2 = 0$ Thus, the equation of the locus is $f(x, y) = x^2 + y^2 + x - 3y - 2 = 0$.

Step 3: Find the points of intersection with the x-axis (A and B). The x-axis is defined by $y=0$. Substitute $y=0$ into the locus equation to find the x-intercepts. $x^2 + 0^2 + x - 3(0) - 2 = 0$ $x^2 + x - 2 = 0$ Factor the quadratic equation: $(x+2)(x-1) = 0$ So, $x = -2$ or $x = 1$. The points of intersection with the x-axis are $A(-2, 0)$ and $B(1, 0)$.

Step 4: Find the points of intersection with the y-axis (C and D). The y-axis is defined by $x=0$. Substitute $x=0$ into the locus equation to find the y-intercepts. $0^2 + y^2 + 0 - 3y - 2 = 0$ $y^2 - 3y - 2 = 0$ Use the quadratic formula to solve for $y$: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-2)}}{2(1)} = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}$ So, the points of intersection with the y-axis are $C\left(0, \frac{3 + \sqrt{17}}{2}\right)$ and $D\left(0, \frac{3 - \sqrt{17}}{2}\right)$.

Step 5: Calculate the lengths of the diagonals of the quadrilateral ACBD. The vertices are $A(-2,0)$, $B(1,0)$, $C\left(0, \frac{3 + \sqrt{17}}{2}\right)$, and $D\left(0, \frac{3 - \sqrt{17}}{2}\right)$. The diagonals are AB and CD, which lie on the x and y axes respectively, and are thus perpendicular. The length of diagonal AB is: $d_1 = |1 - (-2)| = 3$ The length of diagonal CD is: $d_2 = \left|\frac{3 + \sqrt{17}}{2} - \frac{3 - \sqrt{17}}{2}\right| = \left|\frac{2\sqrt{17}}{2}\right| = \sqrt{17}$

Step 6: Calculate the area of the quadrilateral ACBD. The area of quadrilateral ACBD is: $\text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 3 \times \sqrt{17} = \frac{3\sqrt{17}}{2}$

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when expanding and simplifying the equation of the locus.
• Remember the quadratic formula and how to apply it correctly when solving for intercepts.
• Recognize that intercepts on the x and y axes create perpendicular diagonals, simplifying the area calculation.

Summary

We found the locus of point P to be a circle. Then, we determined the x and y intercepts of this circle, which formed the vertices of a quadrilateral with perpendicular diagonals. Finally, we calculated the area of this quadrilateral using the formula $\frac{1}{2}d_1d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.

The final answer is $\boxed{\frac{3\sqrt{17}}{2}}$, which corresponds to option (B).