58. Astronomy Halley's comet has an elliptical orbit with the center of the sun at one focus. The eccentricity of the orbit is approximately 0.967 . The length of the major axis of the orbit is approximately 35.88 astronomical units. (An astronomical unit is about 93 million miles.) (a) Find an equation of the orbit. Place the center of the orbit at the origin and place the major axis on the $$ x $$-axis. (b) Use a graphing utility to graph the equation of the orbit. (c) Find the greatest and least distances (the aphelion and perihelion, respectively) from the sun's center to the comet's center.

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We begin by noting that an ellipse with a horizontal major axis and center at the origin has the equation

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
$$

where $$ a $$ is the semi‐major axis and $$ b $$ is the semi‐minor axis. The foci lie along the $$ x $$-axis at $$(\pm c,0)$$, with

$$
c^2=a^2-b^2.
$$

Because the sun is at one focus and the eccentricity is given by

$$
e=\frac{c}{a},
$$

we can determine $$ c $$, $$ b $$, and then write the equation.

---

**Step 1. Find the value of $$ a $$.**

The length of the major axis is given as 35.88 astronomical units (AU). Hence the semi–major axis is

$$
a=\frac{35.88}{2}=17.94 \text{ AU}.
$$

---

**Step 2. Find the value of $$ c $$ using the eccentricity.**

We are told that

$$
e=0.967=\frac{c}{a}.
$$

Thus,

$$
c=0.967a=0.967(17.94)\approx17.35 \text{ AU}.
$$

---

**Step 3. Find the value of $$ b $$.**

Using the relationship $$ c^2 = a^2 - b^2 $$, we have

$$
b^2=a^2-c^2.
$$

Calculate $$ a^2 $$ and $$ c^2 $$:

$$
a^2=(17.94)^2\approx321.96,
$$
$$
c^2\approx(17.35)^2\approx301.22.
$$

Then

$$
b^2\approx321.96-301.22=20.74,
$$
$$
b\approx\sqrt{20.74}\approx4.55 \text{ AU}.
$$

(Depending on rounding, you may see values like $$4.57$$ AU. Here we take $$b\approx4.55$$ AU.)

---

**Step 4. Write the equation of the ellipse.**

The equation of the orbit, with the center at the origin and the major axis on the $$ x $$-axis, is

$$
\frac{x^2}{(17.94)^2}+\frac{y^2}{(4.55)^2}=1.
$$

That is,

$$
\frac{x^2}{321.96}+\frac{y^2}{20.74}=1.
$$

*Note:* In this coordinate system the sun (which is at one focus) is located at

$$
(c,0)\approx(17.35,0).
$$

---

**Step 5. Graphing the equation.**

To graph the ellipse, you can use a graphing utility and plot

$$
\frac{x^2}{321.96}+\frac{y^2}{20.74}=1.
$$

On the graph:

- The center is at $$(0,0)$$.
- The vertices (endpoints of the major axis) are at $$(\pm 17.94,0)$$.
- The co–vertices (endpoints of the minor axis) are at $$(0, \pm 4.55)$$.
- Mark the sun’s position at $$(17.35,0)$$.

---

**Step 6. Find the aphelion and perihelion distances.**

Recall:

- The **perihelion** is the closest approach of the comet to the sun.
- The **aphelion** is the farthest distance from the sun.

Since the ellipse is centered at the origin with a focus at $$(c,0)$$, the two vertices are at $$ x=a $$ (to the right) and $$ x=-a $$ (to the left). The distance from the sun (at $$(c,0)$$) to a point on the $$ x $$-axis is given by the absolute difference in the $$ x $$-coordinates:

- For the vertex at $$ (a, 0) $$ (i.e. $$ x=+a $$) the distance is

$$
\text{Distance} = a-c\approx17.94-17.35\approx0.59 \text{ AU}.
$$

- For the vertex at $$ (-a, 0) $$ (i.e. $$ x=-a $$) the distance is

$$
\text{Distance} = a+c\approx17.94+17.35\approx35.29 \text{ AU}.
$$

Thus, the perihelion (nearest distance) is approximately $$0.59$$ AU, and the aphelion (farthest distance) is approximately $$35.29$$ AU.

---

**Final Answers:**

(a) The equation of the orbit is

$$
\frac{x^2}{321.96}+\frac{y^2}{20.74}=1 \quad \text{or equivalently} \quad \frac{x^2}{(17.94)^2}+\frac{y^2}{(4.55)^2}=1.
$$

(b) When graphing the ellipse, plot the above equation with center $$(0,0)$$, vertices at $$(\pm 17.94,0)$$, co–vertices at $$(0,\pm 4.55)$$, and mark the sun’s position at $$(17.35,0)$$.

(c) The distances from the sun to the comet are approximately:
- Perihelion: $$ a-c\approx0.59 $$ AU,
- Aphelion: $$ a+c\approx35.29 $$ AU.
(a) The equation of the orbit is $$ \frac{x^2}{321.96} + \frac{y^2}{20.74} = 1 $$. (b) Graph the ellipse with center at (0,0), vertices at (±17.94,0), co-vertices at (0,±4.55), and mark the sun at (17.35,0). (c) The perihelion is approximately 0.59 AU and the aphelion is approximately 35.29 AU.

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