Ellipses: Additional Practice (Set 1)

The denominators are equal, so $a^2 = b^2 = 36$ and the focal distance vanishes.

$c^2 = 36 - 36 = 0 \quad\Rightarrow\quad c = 0 \quad\Rightarrow\quad e = \frac{c}{a} = 0$

Both foci sit at the center, and the curve is the circle $x^2 + y^2 = 36$ of radius $6$. Eccentricity $0$ is exactly the circle case, and no ellipse can reach $e = 1$.

The base is the major axis, so $2a = 50$ and $a = 25$; the central height is the semi-minor axis, so $b = 20$. The arch is the top half of $\frac{x^2}{625} + \frac{y^2}{400} = 1$. Put $x = 15$:

$\frac{225}{625} + \frac{y^2}{400} = 1 \quad\Rightarrow\quad \frac{y^2}{400} = 1 - \frac{9}{25} = \frac{16}{25}$

Multiply by $400$ and take the positive root:

$y^2 = 256 \quad\Rightarrow\quad y = 16$

The arch is $16$ feet high there, still most of its full height, because an ellipse stays high near its center and drops steeply near the ends.

Foci on the $y$-axis means the major axis is vertical, so $16$ must be the larger denominator: $a^2 = 16$ and $b^2 = m < 16$. The separation of the foci is $2c$, so

$2c = 2\sqrt{7} \quad\Rightarrow\quad c = \sqrt{7} \quad\Rightarrow\quad c^2 = 7.$

Now solve $c^2 = a^2 - b^2$ for $m$:

$7 = 16 - m \quad\Rightarrow\quad m = 9$

Since $9 < 16$, the major axis really is vertical. The value $m = 23$ also gives $c^2 = 23 - 16 = 7$, but then $m$ is the larger denominator and the foci would land on the $x$-axis, so it fails the stated condition.

The sun sits at a focus, which is $c$ from the center, and $e = \frac{c}{a}$ gives $c = ea$.

$c = 0.017 \cdot 150 \approx 2.6 \text{ million km}$

Compared with the $150$ million km semi-major axis, that offset is tiny, which is why Earth's orbit looks like a circle in any drawing. It is the position of the sun, not the shape of the curve, that gives the eccentricity away.

The ellipse touches the left and right sides at its vertices and the top and bottom at its co-vertices, so the rectangle's width is $2a$ and its height is $2b$.

$2a = 26 \Rightarrow a = 13, \qquad 2b = 10 \Rightarrow b = 5$

The major axis is horizontal, since $a > b$, so the foci are on the $x$-axis.

$c^2 = 169 - 25 = 144 \quad\Rightarrow\quad c = 12$

The foci are $(\pm 12, 0)$. The points $(\pm 13, 0)$ are the vertices.

Substitute $y = x$ into the ellipse and combine the terms.

$\frac{x^2}{4} + x^2 = 1 \quad\Rightarrow\quad \frac{5x^2}{4} = 1 \quad\Rightarrow\quad x^2 = \frac{4}{5}$

Take roots and rationalize the denominator:

$x = \pm \frac{2}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5}$

Since $y = x$, the two intersection points are $\left(\frac{2\sqrt{5}}{5}, \frac{2\sqrt{5}}{5}\right)$ and its negative. The point $(1, 1)$ fails, because $\frac{1}{4} + 1 = \frac{5}{4} \ne 1$.

The vertices of $\frac{x^2}{25} + \frac{y^2}{9} = 1$ are $(\pm 5, 0)$. Sharing them fixes the center (their midpoint, the origin), fixes the major axis as the horizontal segment between them, and gives $a = 5$, so only $b$ is free to change. Turn the new eccentricity into a value of $c$:

$c = ea = \frac{3}{5} \cdot 5 = 3$

Then $b^2 = a^2 - c^2 = 25 - 9 = 16$, giving $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

The original ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ has $c = \sqrt{25 - 9} = 4$, hence $e = \frac{4}{5}$, so it is not the answer. The ellipse $\frac{x^2}{16} + \frac{y^2}{25} = 1$ does have eccentricity $\frac{3}{5}$, but its vertices are $(0, \pm 5)$, on the wrong axis.

Divide by $36$ to reach standard form.

$\frac{x^2}{9} + \frac{y^2}{4} = 1$

The larger denominator gives $a^2 = 9$, so $a = 3$. The constant focal sum is the string length, which is the full major axis:

$r_1 + r_2 = 2a = 6$

The number $3$ is only the semi-major axis, half of what the question asks for.

The point $(6, -2)$ sits on the horizontal line through the center, $5$ units to the right, so it is a vertex and $a = 5$. The point $(1, 1)$ sits directly above the center, $3$ units up, so it is a co-vertex and $b = 3$.

$c^2 = a^2 - b^2 = 25 - 9 = 16 \quad\Rightarrow\quad c = 4$

The foci are $4$ units left and right of $(1, -2)$, at $(5, -2)$ and $(-3, -2)$. Adding the squares instead would give the impossible $\sqrt{34} > a$.

Roundness is measured by eccentricity, so compute $e = \frac{c}{a}$ for each curve, taking $c^2$ to be the larger denominator minus the smaller one.

$\frac{x^2}{26} + \frac{y^2}{25} = 1 \;\Rightarrow\; e = \frac{1}{\sqrt{26}} \approx 0.20, \qquad \frac{x^2}{25} + \frac{y^2}{16} = 1 \;\Rightarrow\; e = \frac{3}{5} = 0.6$

$\frac{x^2}{100} + \frac{y^2}{36} = 1 \;\Rightarrow\; e = \frac{8}{10} = 0.8, \qquad \frac{x^2}{9} + \frac{y^2}{4} = 1 \;\Rightarrow\; e = \frac{\sqrt{5}}{3} \approx 0.75$

The smallest eccentricity belongs to $\frac{x^2}{26} + \frac{y^2}{25} = 1$, whose two denominators are nearly equal. Equal denominators would give $c = 0$ and $e = 0$, a circle exactly, so denominators this close give a nearly circular ellipse.

The distance from a point of the ellipse to a focus runs from $a - c$ at the near vertex to $a + c$ at the far one, so the two given distances are exactly those extremes.

$(a - c) + (a + c) = 2a = 363{,}000 + 405{,}000 = 768{,}000$

The $c$ terms cancel, so

$a = \frac{768{,}000}{2} = 384{,}000 \text{ km}.$

Averaging the closest and farthest distances gives the semi-major axis, which is why $a$ is often quoted as the moon's mean distance.

Subtracting the two extremes isolates $c$, since $(a + c) - (a - c) = 2c$.

$2c = 405{,}000 - 363{,}000 = 42{,}000 \quad\Rightarrow\quad c = 21{,}000$

Adding them gave $2a = 768{,}000$, so $a = 384{,}000$. Now divide:

$e = \frac{c}{a} = \frac{21{,}000}{384{,}000} \approx 0.055$

That is a nearly circular orbit, and it matches the measured value of about $0.055$.