Question

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

• A. (0, \( \pm \)3)
• B. (3, \( \pm \)3)
• C. (0, \( \pm \)5)
• D. (5, \( \pm \)5)

Hint:

For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b>a$, foci are $(0, \pm be)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the coordinates of the foci of a given conic section. We first need to identify the type of conic and its standard form.
Step 2: Key Formula or Approach:
1. Rearrange the given equation into the standard form of an ellipse, \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) (for a vertical ellipse) or \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) (for a horizontal ellipse), by completing the square.
2. Identify the center (h, k), and the values of \(a^2\) and \(b^2\).
3. Calculate the distance from the center to the foci, \(c\), using the formula \(c^2 = a^2 - b^2\).
4. Determine the coordinates of the foci, which are (h, k \( \pm \) c) for a vertical ellipse or (h \( \pm \) c, k) for a horizontal ellipse.
Step 3: Detailed Explanation:
The given equation is \( 25x^2 + 16y^2 - 150x = 175 \).
1. Complete the square:
Group the x-terms and y-terms:
\[ (25x^2 - 150x) + 16y^2 = 175 \] Factor out the coefficient of the squared term from the x-group:
\[ 25(x^2 - 6x) + 16y^2 = 175 \] To complete the square for \(x^2 - 6x\), we take half of the coefficient of x (-6), which is -3, and square it to get 9. We add and subtract 9 inside the parenthesis:
\[ 25(x^2 - 6x + 9 - 9) + 16y^2 = 175 \] \[ 25((x - 3)^2 - 9) + 16y^2 = 175 \] Distribute the 25:
\[ 25(x - 3)^2 - 25(9) + 16y^2 = 175 \] \[ 25(x - 3)^2 - 225 + 16y^2 = 175 \] Move the constant term to the right side:
\[ 25(x - 3)^2 + 16y^2 = 175 + 225 \] \[ 25(x - 3)^2 + 16y^2 = 400 \] Divide the entire equation by 400 to get 1 on the right side:
\[ \frac{25(x - 3)^2}{400} + \frac{16y^2}{400} = 1 \] \[ \frac{(x - 3)^2}{16} + \frac{y^2}{25} = 1 \] 2. Identify parameters:
This is the standard form of a vertical ellipse because the denominator under the \(y^2\) term (25) is larger than the denominator under the \(x^2\) term (16).
- Center (h, k) = (3, 0).
- \(a^2 = 25 \implies a = 5\) (major radius).
- \(b^2 = 16 \implies b = 4\) (minor radius).
3. Calculate c:
\[ c^2 = a^2 - b^2 = 25 - 16 = 9 \] \[ c = \sqrt{9} = 3 \] 4. Find the foci:
For a vertical ellipse, the foci are located at (h, k \( \pm \) c).
Foci = (3, 0 \( \pm \) 3).
The coordinates of the two foci are (3, 3) and (3, -3). This can be written as (3, \( \pm \)3).
Step 4: Final Answer:
The foci of the conic are at (3, \( \pm \)3), which matches option (B).