The equation \(\frac{(\mathrm{x} - 5)^2}{16} + \frac{(\mathrm{y} + 1)^2}{9} = 1\) represents an ellipse in the xy-plane. A point \((\mathrm{p},...

1. TRANSLATE the equation into standard form

• Given equation: \(\frac{(\mathrm{x} - 5)^2}{16} + \frac{(\mathrm{y} + 1)^2}{9} = 1\)
• This matches standard ellipse form: \(\frac{(\mathrm{x} - \mathrm{h})^2}{\mathrm{a}^2} + \frac{(\mathrm{y} - \mathrm{k})^2}{\mathrm{b}^2} = 1\)
• What this tells us:
  • Center: \((\mathrm{h}, \mathrm{k}) = (5, -1)\)
  • \(\mathrm{a}^2 = 16\), so \(\mathrm{a} = 4\)
  • \(\mathrm{b}^2 = 9\), so \(\mathrm{b} = 3\)

2. INFER the constraints on x-coordinates

• Since \(\mathrm{a} = 4\) is the semi-major axis length (horizontal direction), any point \((\mathrm{p}, \mathrm{q})\) on this ellipse must have its x-coordinate within 4 units of the center's x-coordinate
• The valid range for \(\mathrm{p}\) is: \(\mathrm{center}_x - \mathrm{a} \leq \mathrm{p} \leq \mathrm{center}_x + \mathrm{a}\)

3. SIMPLIFY to find the exact range

• Substituting our values: \(5 - 4 \leq \mathrm{p} \leq 5 + 4\)
• Therefore: \(1 \leq \mathrm{p} \leq 9\)

4. APPLY CONSTRAINTS to check answer choices

• (A) -1: Outside the range \([1, 9]\) ✗
• (B) 0: Outside the range \([1, 9]\) ✗
• (C) 3: Within the range \([1, 9]\) ✓
• (D) 11: Outside the range \([1, 9]\) ✗

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students don't recognize the standard form or incorrectly identify the center and axis lengths.

For example, they might think the center is at (0, 0) or confuse which values represent a and b. This leads to calculating the wrong range for possible x-coordinates, causing them to incorrectly eliminate valid choices or accept invalid ones. This may lead them to select Choice A (-1) or Choice D (11) if they calculate an incorrect range.

Second Most Common Error:

Poor INFER reasoning: Students understand the ellipse parameters but don't connect this to the constraint that x-coordinates must lie within the horizontal span of the ellipse.

They might attempt to substitute each answer choice directly into the equation rather than using the geometric constraint. While this approach could work, it's more time-consuming and prone to calculation errors. This leads to confusion and potentially guessing among the choices.

The Bottom Line:

This problem tests whether students can efficiently use the geometric properties of an ellipse rather than getting bogged down in algebraic manipulation. The key insight is that the standard form immediately gives you the boundaries for all possible coordinate values.