<p>If <span class="math">\(S\)</span> and <span class="math">\(S'\)</span> are the foci of the ellipse <span class="math">\(\frac{x^2}{25} + \frac{y^2}{16} = 1\)</span> and <span class="math">\(P\)</span> is any point on it, then the difference of maximum and minimum of <span class="math">\(SP \cdot S'P\)</span> is equal to:</p>
<p>(a) <span class="math">\(16\)</span></p>
<p>(b) <span class="math">\(9\)</span></p>
<p>(c) <span class="math">\(15\)</span></p>
<p>(d) <span class="math">\(25\)</span></p>
Step-by-Step Solution
Key Concept: For any point P on an ellipse with foci S and S', we have SP + S'P = 2a (constant). Using this constraint along with the AM-GM inequality and distance formulas, we can find the extrema of the product SP·S'P.
<p><strong>Step 1:</strong> Identify ellipse parameters. From $\frac{x^2}{25} + \frac{y^2}{16} = 1$, we have $a^2 = 25$ and $b^2 = 16$, so $a = 5$ and $b = 4$.</p><p><strong>Step 2:</strong> Find eccentricity and foci. $c^2 = a^2 - b^2 = 25 - 16 = 9$, so $c = 3$. Foci are at $S = (-3, 0)$ and $S' = (3, 0)$.</p><p><strong>Step 3:</strong> Use the fundamental ellipse property. For any point P on the ellipse: $SP + S'P = 2a = 10$ (constant).</p><p><strong>Step 4:</strong> Let $SP = r_1$ and $S'P = r_2$. We have $r_1 + r_2 = 10$. We need to find max and min of $r_1 \cdot r_2$ subject to this constraint.</p><p><strong>Step 5:</strong> By AM-GM inequality: $\frac{r_1 + r_2}{2} \geq \sqrt{r_1 \cdot r_2}$, which gives $5 \geq \sqrt{r_1 \cdot r_2}$, so $r_1 \cdot r_2 \leq 25$. Equality holds when $r_1 = r_2 = 5$, at the minor axis endpoints $(0, \pm 4)$. Maximum is $25$.</p><p><strong>Step 6:</strong> For minimum, consider the major axis endpoints. At $(5, 0)$: $SP = 8, S'P = 2$, so $SP \cdot S'P = 16$. At $(-5, 0)$: $SP = 2, S'P = 8$, so $SP \cdot S'P = 16$. Minimum is $16$.</p><p><strong>Step 7:</strong> Difference = Maximum - Minimum = $25 - 16 = 9$.</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A