Question

find the ends of the major axis and foci of this ellipse.

49x² + 16y² = 784

major axis : (0, ±?)

foci : (0, ±√ )

Explanation:

Step1: Write the equation in standard form

Divide the given equation \(49x^{2}+16y^{2}=784\) by \(784\).
We get \(\frac{x^{2}}{16}+\frac{y^{2}}{49}=1\).
For an ellipse of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (\(a>b>0\)), the major - axis endpoints are \((0,\pm a)\) and \(c=\sqrt{a^{2}-b^{2}}\) (where \(c\) is the distance from the center to the foci).
Here, \(a^{2}=49\), so \(a = 7\).

Step2: Calculate \(c\)

Since \(b^{2}=16\) and \(a^{2}=49\), then \(c=\sqrt{a^{2}-b^{2}}=\sqrt{49 - 16}\).
\(c=\sqrt{33}\).

Answer:

Major axis: \((0,\pm7)\)
Foci: \((0,\pm\sqrt{33})\)