Digital ACT Math Practice Question #1622
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c = sqrt(25 - 16) = 3. Distance between TWO foci is 2c = 6.
For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a > b\), the foci lie on the major axis at distance \(c\) from the center, where \(c^2 = a^2 - b^2\). Here, \(a^2 = 25\) and \(b^2 = 16\), so \(c^2 = 25 - 16 = 9 \implies c = 3...
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