Question

The foci of the conic section \(25x^2 + 16y^2 - 150x - 175 = 0\) are

• A. \((0, \pm 3)\)
• B. \((0, \pm 2)\)
• C. \((3, \pm 3)\)
• D. \((0, \pm 1)\)

Hint:

For ellipse \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) with \(a > b\), foci are at \((h, k \pm c)\) where \(c^2 = a^2 - b^2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Complete the square to identify the conic.

Step 2: Detailed Explanation:
\(25x^2 - 150x + 16y^2 = 175\)
\(25(x^2 - 6x) + 16y^2 = 175\)
\(25(x^2 - 6x + 9) + 16y^2 = 175 + 225\)
\(25(x - 3)^2 + 16y^2 = 400\)
\(\frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1\)
Ellipse: center \((3, 0)\), \(a^2 = 25\), \(b^2 = 16\), so \(c^2 = a^2 - b^2 = 9\), \(c = 3\).
Foci: \((3, \pm 3)\).

Step 3: Final Answer:
Option (C) \((3, \pm 3)\).