Question

what are the foci of the ellipse $x^{2}+4y^{2}-12x - 64 = 0$? write your answer in simplified, rationalized form. and

Explanation:

Step1: Rewrite the equation in standard form

Group \(x\) - terms and complete the square:

$$
LATEXBLOCK0
$$

For an ellipse of the form \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}}=1\) (\(a>b>0\)), here \(h = 6,k = 0,a^{2}=100\), so \(a = 10\), \(b^{2}=25\), so \(b = 5\).

Step2: Calculate \(c\)

Use the relationship \(c^{2}=a^{2}-b^{2}\)

$$ c^{2}=100 - 25=75 $$

So \(c=\sqrt{75}=5\sqrt{3}\)

Step3: Find the foci

Since the major - axis is parallel to the \(x\) - axis (because \(a^{2}\) is under the \(x\) - term), the foci are \((h\pm c,k)\)
Substitute \(h = 6,k = 0,c = 5\sqrt{3}\)

Answer:

\((6 + 5\sqrt{3},0)\) and \((6-5\sqrt{3},0)\)