Introduction: The Defining Property and Focal Property of an Ellipse

An ellipse is a fundamental conic section defined by a unique geometric property: it is the locus of a point P such that the sum of its distances from two fixed points, called foci ($F_1$ and $F_2$), is a constant value. This constant sum is equal to the length of the major axis, which is denoted as $2a$.

For an ellipse centered at the origin $(0,0)$ with its major axis lying along the x-axis (meaning $a > b$), its standard equation is: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ The coordinates of its foci are $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$, where $a$ is the length of the semi-major axis, and $e$ is the eccentricity of the ellipse. The eccentricity is a measure of how "stretched out" the ellipse is, and it's calculated using the relation $b^2 = a^2(1 - e^2)$, or equivalently, $e = \sqrt{1 - \frac{b^2}{a^2}}$.

The fundamental focal property states that for any point P on the ellipse, and its two foci $F_1$ and $F_2$: $PF_1 + PF_2 = 2a$ This property is the cornerstone for solving this problem.

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Step-by-Step Solution:

1. Standardize the Equation of the Given Conic

The equation of the conic is given as $9x^2 + 16y^2 = 144$. Our first task is to transform this equation into the standard form of an ellipse, which is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This standardization is crucial because it allows us to directly identify the values of $a^2$ and $b^2$, which are essential for determining the ellipse's dimensions and orientation.

To achieve this, we divide the entire equation by the constant term on the right-hand side, which is 144: $\frac{9x^2}{144} + \frac{16y^2}{144} = \frac{144}{144}$ Simplifying the fractions, we get: $\frac{x^2}{16} + \frac{y^2}{9} = 1$ Explanation: This step is foundational. By obtaining '1' on the right-hand side, we can now clearly see the denominators that correspond to $a^2$ and $b^2$. The positive coefficients for both $x^2$ and $y^2$ and their different denominators confirm that this conic is indeed an ellipse.

2. Identify the Semi-Major and Semi-Minor Axes of the Ellipse

Now we compare our standardized equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$ with the general standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

From this comparison, we can directly identify the squared lengths of the semi-axes:

• $a^2 = 16 \implies a = \sqrt{16} = 4$ (since $a$ represents a length, it must be positive)
• $b^2 = 9 \implies b = \sqrt{9} = 3$ (similarly, $b$ must be positive)

Explanation: The values $a$ and $b$ represent the lengths of the semi-major and semi-minor axes, respectively. Since $a = 4$ and $b = 3$, we observe that $a > b$. This condition ($a>b$) is significant: it tells us that the major axis of this ellipse lies along the x-axis, and its center is at the origin $(0,0)$. This orientation is critical for correctly determining the coordinates of the foci in the next step.

3. Determine the Foci of the Ellipse

To locate the foci of the ellipse, we first need to calculate its eccentricity, $e$. For an ellipse with its major axis along the x-axis (as determined by $a>b$), the formula for eccentricity is: $e = \sqrt{1 - \frac{b^2}{a^2}}$ Substitute the values of $a^2 = 16$ and $b^2 = 9$ into the formula: $e = \sqrt{1 - \frac{9}{16}}$ To simplify the expression inside the square root, find a common denominator: $e = \sqrt{\frac{16 - 9}{16}}$ $e = \sqrt{\frac{7}{16}}$ $e = \frac{\sqrt{7}}{4}$ Now that we have the eccentricity, we can find the coordinates of the foci. For an ellipse with its major axis along the x-axis, the foci are located at $(\pm ae, 0)$. Let's calculate the value of $ae$: $ae = 4 \times \frac{\sqrt{7}}{4} = \sqrt{7}$ Therefore, the coordinates of the foci of this ellipse are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$.

Explanation: The eccentricity $e$ is a key parameter that quantifies the "roundness" or "elongation" of an ellipse. It helps us pinpoint the exact locations of the foci, which are at a distance of $ae$ from the center. Calculating $ae$ is a crucial step in preparing to apply the focal property.

4. Connect the Given Points A and B to the Ellipse's Foci

The problem statement provides the coordinates of two points A and B as $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$ respectively.

Let's compare these given coordinates with the foci we calculated in Step 3, which are $(\sqrt{7}, 0)$ and $(-\sqrt{7}, 0)$. We can see that there is a perfect match. This means that points A and B are precisely the foci of the given ellipse.

Explanation: This is a pivotal moment in solving the problem. The question is cleverly designed to test your ability to recognize that the given points A and B are actually the foci of the ellipse. If you make this connection, the problem simplifies dramatically, allowing for the direct application of the ellipse's defining property. If you missed this, you might attempt to use the distance formula, which would be much more complicated and prone to errors.

5. Apply the Focal Property of the Ellipse to Find PA + PB

According to the fundamental definition of an ellipse (its focal property), for any point P on the ellipse, the sum of its distances from the two foci is constant and equal to the length of the major axis, $2a$.

Since P is any point on the ellipse, and we have established that A and B are its foci, we can directly apply this property: $PA + PB = 2a$ From Step 2, we found that the length of the semi-major axis is $a = 4$. Substitute this value into the equation: $PA + PB = 2 \times 4$ $PA + PB = 8$

Explanation: This step directly utilizes the core concept of an ellipse's definition. Once it's confirmed that A and B are the foci, the problem becomes a straightforward application of this definition. The sum of distances from any point on the ellipse to its foci is a constant value, which is always equal to the length of the major axis ($2a$).

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Expert Tips and Common Pitfalls:

• Standardizing the Equation: Always begin by ensuring the right-hand side of the conic equation is 1. If it's not, divide the entire equation by that constant. This is a common first step for almost all conic problems.
• Identifying Major/Minor Axes Correctly: For an ellipse equation $\frac{x^2}{D_1} + \frac{y^2}{D_2} = 1$:
  • If $D_1 > D_2$, then $a^2 = D_1$ and $b^2 = D_2$. The major axis is along the x-axis, and foci are $(\pm ae, 0)$.
  • If $D_2 > D_1$, then $a^2 = D_2$ and $b^2 = D_1$. The major axis is along the y-axis, and foci are $(0, \pm ae)$.
  • In this problem, $16 > 9$, so $a^2=16$ (under $x^2$) implies the major axis is along the x-axis. A common mistake is to always assume $a^2$ is under $x^2$.
• Eccentricity Formula: Remember the correct form for eccentricity: $e = \sqrt{1 - \frac{\text{square of semi-minor axis}}{\text{square of semi-major axis}}}$. This ensures you use $\frac{b^2}{a^2}$ when $a$ is the semi-major axis and $\frac{a^2}{b^2}$ when $b$ is the semi-major axis.
• Focal Property Value: The constant sum of distances from any point on the ellipse to its foci is always equal to the length of the major axis, $2a$. It is never $2b$ (the length of the minor axis).
• Read Carefully and Connect the Dots: The problem's phrasing "A and B are... and P is any point on the conic" is a strong hint to look for the conic's definition if A and B turn out to be the foci. Always check this possibility after finding the foci. It saves a lot of calculation!

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Summary and Key Takeaway:

This problem is an excellent test of your foundational understanding of the geometric definition and properties of an ellipse. The efficiency and correctness of the solution hinge on a few key steps:

1. Standardizing the Conic Equation: Convert $9x^2 + 16y^2 = 144$ into the standard form $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
2. Identifying Semi-Axes: From the standard form, determine $a=4$ and $b=3$, noting that $a>b$ means the major axis is along the x-axis.
3. Calculating Foci: Calculate the eccentricity $e = \frac{\sqrt{7}}{4}$ and then the focal coordinates $(\pm ae, 0) = (\pm \sqrt{7}, 0)$.
4. Crucial Recognition: Observe that the given points A and B are precisely the foci of the ellipse you just analyzed.
5. Applying the Defining Property: Since A and B are the foci and P is on the ellipse, the sum of distances $PA + PB$ is equal to the length of the major axis, $2a$.

By following these steps, we find that $PA + PB = 2a = 2 \times 4 = 8$. This problem elegantly demonstrates how understanding the basic definition of a conic section can lead to a quick and straightforward solution.

The final answer is $\boxed{\text{8}}$.