Question

81x² + 16y² = 1296
choose the correct graph below.
a. graph of an ellipse centered at origin, vertical major axis
b. graph of a narrow ellipse centered at origin, vertical major axis
c. graph of an ellipse centered at origin, horizontal major axis
locate the foci.
(type ordered pairs. use a comma to separate answers. type exact answers, using radicals as needed. simplify your answer)

Explanation:

Step1: Rewrite the equation in standard form

The general equation of an ellipse is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (for vertical major axis) or \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (for horizontal major axis), where \(a>b>0\). We start with \(81x^{2}+16y^{2}=1296\). Divide both sides by \(1296\):

$$ \frac{81x^{2}}{1296}+\frac{16y^{2}}{1296}=1 $$

Simplify the fractions: \(\frac{x^{2}}{16}+\frac{y^{2}}{81}=1\). Here, \(a^{2} = 81\) (so \(a = 9\)) and \(b^{2}=16\) (so \(b = 4\)). Since \(a^{2}\) is under the \(y^{2}\) term, the major axis is vertical.

Step2: Find the distance from center to foci (\(c\))

For an ellipse, the relationship is \(c^{2}=a^{2}-b^{2}\). Substitute \(a^{2}=81\) and \(b^{2}=16\):

$$ c^{2}=81 - 16=65 $$

So \(c=\sqrt{65}\). The center of the ellipse is at \((0,0)\) (since the equation is \(\frac{x^{2}}{16}+\frac{y^{2}}{81}=1\), no shifts in \(x\) or \(y\)). For a vertical major axis, the foci are at \((0,\pm c)\), so the foci are \((0,\sqrt{65})\) and \((0,-\sqrt{65})\).

Step3: Identify the correct graph

Since the major axis is vertical (longer along the \(y\)-axis), the ellipse should be taller than it is wide. Graph A has a vertical major axis (taller ellipse), Graph B is a very narrow vertical ellipse (but our \(a = 9\), \(b = 4\), so the width is \(2b=8\) and height is \(2a = 18\), so it should be a more "balanced" tall ellipse), and Graph C has a horizontal major axis. So the correct graph is A (as it's already marked, and matches the vertical major axis with appropriate proportions).

Answer:

The correct graph is A. The foci are \((0,\sqrt{65}),(0,-\sqrt{65})\)