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Question 1209671: Square ABCD has side length 6. An ellipse $\mathcal{E}$ is circumscribed about the square and there is a point $P$ on the ellipse such that $PC = PD = 8$. What is the area of ellipse $\mathcal{E}$? (You may assume the sides of the square are parallel to the axes of the ellipse.)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
1. **Coordinate System:** Place the square in a coordinate system so that its center is at the origin (0,0). Since the side length is 6, the vertices of the square are at (±3, ±3).
2. **Ellipse Equation:** The general equation of an ellipse centered at the origin is (x²/a²) + (y²/b²) = 1, where 'a' is the semi-major axis and 'b' is the semi-minor axis.
3. **Vertices on the Ellipse:** The vertices of the square lie on the ellipse. Therefore, the points (±3, ±3) must satisfy the ellipse equation:
(3²/a²) + (3²/b²) = 1
9/a² + 9/b² = 1
4. **Point P:** The point P is such that PC = PD = 8. Let the coordinates of P be (x, y). Using the distance formula:
PC² = (x - 3)² + (y - 3)² = 8² = 64
PD² = (x + 3)² + (y - 3)² = 8² = 64
5. **Solving for x and y:** Subtract the two equations:
[(x - 3)² - (x + 3)²] + [(y - 3)² - (y - 3)²] = 0
(x² - 6x + 9 - x² - 6x - 9) = 0
-12x = 0
x = 0
Now substitute x = 0 into the equation for PC²:
(0 - 3)² + (y - 3)² = 64
9 + (y - 3)² = 64
(y - 3)² = 55
y - 3 = ±√55
y = 3 ± √55
So, the possible coordinates for P are (0, 3 + √55) and (0, 3 - √55). Since the ellipse is circumscribed around the square, the value 3 - √55 must be between -3 and 3. √55 is approximately 7.4, so 3 - √55 ≈ -4.4, which is outside the range. The y coordinate we want is 3 + √55.
6. **Substituting P into the Ellipse Equation:** Since P(0, 3 + √55) lies on the ellipse:
(0²/a²) + ( (3 + √55)² / b² ) = 1
(9 + 6√55 + 55) / b² = 1
(64 + 6√55) / b² = 1
b² = 64 + 6√55
7. **Solving for a²:** Substitute b² back into the equation from step 3:
9/a² + 9/(64 + 6√55) = 1
9/a² = 1 - 9/(64 + 6√55)
9/a² = (64 + 6√55 - 9) / (64 + 6√55)
9/a² = (55 + 6√55) / (64 + 6√55)
a² = 9 * (64 + 6√55) / (55 + 6√55)
a² = 9 * (64 + 6√55) * (55 - 6√55) / (55² - (6√55)²)
a² = 9 * (3520 - 396√55 + 330√55 - 1980) / (3025 - 1980)
a² = 9 * 1540 / 1045
a² = 9 * (28/19) = 252/19
8. **Area of the Ellipse:** Area = πab = π√(a²b²)
Area = π√[ (252/19) * (64 + 6√55) ]
Area = π√[ (252/19) * (64 + 6√55) ] ≈ π√(252/19 * 108.7) ≈ π√1440 ≈ 12π√10 ≈ 119.38
Final Answer: The final answer is $\boxed{12\pi\sqrt{10}}$
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