To find the foci of the conic section given by the equation \(25x^2 + 16y^2 - 150x - 175 = 0\), we need to first determine the standard form of the conic. Let's solve the problem step-by-step:
- Rearrange the equation: The given equation is \(25x^2 + 16y^2 - 150x - 175 = 0\). Let's rearrange this to make it easier to convert to the standard form.
- Move the constants to the right side: \(25x^2 - 150x + 16y^2 = 175\)
- Complete the square: We will complete the square for both \(x\) and \(y\) terms.
- For the \(x\)-terms: \(25(x^2 - 6x)\)
Take \(-6\), halve it, and square it to complete the square:
\((-6/2)^2 = 9\)
So: \(25(x^2 - 6x + 9 - 9)\)
= \(25((x-3)^2 - 9)\)
= \(25(x-3)^2 - 225\) - For the \(y\)-terms: \(16y^2\) is already a complete square.
- Substitute back to the equation:
- Substitute back the completed square terms:
\(25(x-3)^2 - 225 + 16y^2 = 175\)
\(25(x-3)^2 + 16y^2 = 175 + 225\)
\(25(x-3)^2 + 16y^2 = 400\)
- Divide through by 400 to set into standard form:
- Divide the entire equation by 400:
\(\frac{25(x-3)^2}{400} + \frac{16y^2}{400} = 1\)
\(\frac{(x-3)^2}{16} + \frac{y^2}{25} = 1\)
This is the equation of an ellipse in standard form with center \((3, 0)\).
- Identify the foci of the ellipse:
- The standard form of an ellipse is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
Here, \(a^2 = 16\) and \(b^2 = 25\),
\(a = 4\) and \(b = 5\), and hence \(b > a\). - For an ellipse with vertical major axis, focus is calculated using \(c = \sqrt{b^2 - a^2}\).
\(c = \sqrt{25 - 16} = \sqrt{9} = 3\). - The foci are at: \((h, k \pm c) = (3, 0 \pm 3)\) or \((3, \pm 3)\).
Therefore, the foci of the conic section are \((3, \pm 3)\).