To find Pluto's farthest distance from the sun, we start with the equation for an ellipse in the standard form:
\[ \frac{(x + 0.9)^2}{12.96} + \frac{y^2}{12.15} = 1 \]
Here, we can identify \(a^2\) and \(b^2\):
- The semi-major axis \(a^2 = 12.96\)
- The semi-minor axis \(b^2 = 12.15\)
Now, we calculate \(a\) and \(b\):
\[ a = \sqrt{12.96} \approx 3.6 \quad \text{(in billions of miles)} \] \[ b = \sqrt{12.15} \approx 3.48 \quad \text{(in billions of miles)} \]
Next, we check the placement of the foci using the formula:
\[ c = \sqrt{a^2 - b^2} \]
Calculating \(c\):
\[ c = \sqrt{12.96 - 12.15} = \sqrt{0.81} = 0.9 \quad \text{(in billions of miles)} \]
The foci of the ellipse are located at \((\pm c, 0)\). Since the sun is at one focus, its coordinates are \((0, 0)\), and the distance from the center of the ellipse to a focus is \(c = 0.9\).
The center of the ellipse can be found at \((-0.9, 0)\). The farthest distance from the sun (located at \((0, 0)\)) to Pluto occurs at the point on the ellipse directly opposite the sun along the major axis.
Distance from the center to the farthest point (which is at \((-0.9 + a, 0)\)) is:
\[ \text{Farthest distance} = |c| + a \] Substituting the values:
\[ \text{Farthest distance} = 0.9 + 3.6 = 4.5 \text{ billion miles} \]
Thus, Pluto's farthest distance from the sun is:
4.50 billion mi.