Question

$\frac{x^2}{49} + \frac{y^2}{36} = 1$ how far from the center are the foci located? $\sqrt{13}$ 1 $\sqrt{85}$

Explanation:

Step1: Identify the ellipse standard form

The given equation is \(\frac{x^{2}}{49}+\frac{y^{2}}{36} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (where \(a>b\) for a horizontal major axis). Here, \(a^{2} = 49\) so \(a = 7\), and \(b^{2}=36\) so \(b = 6\).

Step2: Use the ellipse foci formula

For an ellipse, the relationship between \(a\), \(b\), and \(c\) (distance from center to foci) is \(c^{2}=a^{2}-b^{2}\). Substituting the values, we get \(c^{2}=49 - 36=13\). Then, taking the square root, \(c=\sqrt{13}\).

Answer:

\(\sqrt{13}\)