Ellipses: Free Response

Compare the two denominators first, since the larger one names $a^2$.

Here $289 > 64$ and $289$ sits under $y^2$, so the major axis is vertical and

$a^2 = 289, \qquad b^2 = 64 \quad\Rightarrow\quad a = 17, \; b = 8.$

Find $c$ by subtracting, never adding:

$c^2 = a^2 - b^2 = 289 - 64 = 225 \quad\Rightarrow\quad c = 15.$

With the major axis vertical, the vertices and foci lie on the $y$-axis, $a$ and $c$ from the center, and the co-vertices lie on the $x$-axis, $b$ from the center: vertices $(0, \pm 17)$, co-vertices $(\pm 8, 0)$, foci $(0, \pm 15)$.

Now the larger denominator sits under $x^2$: $625 > 49$, so this ellipse has a horizontal major axis, and

$a^2 = 625, \qquad b^2 = 49 \quad\Rightarrow\quad a = 25, \; b = 7.$

Subtract to find $c$:

$c^2 = a^2 - b^2 = 625 - 49 = 576 \quad\Rightarrow\quad c = 24.$

With the major axis horizontal, the vertices and foci lie on the $x$-axis and the co-vertices lie on the $y$-axis: vertices $(\pm 25, 0)$, co-vertices $(0, \pm 7)$, foci $(\pm 24, 0)$.

Treat $a$, $b$ as unspecified positive numbers and $c$ as an unspecified nonnegative number, linked by $b^2 = a^2 - c^2$, and see what that single equation forces, without plugging in either part's numbers.

Rewrite it as

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

Since $c^2 \ge 0$ for any real $c$, adding it to $b^2$ can only make the total at least as large as $b^2$ alone:

$a^2 = b^2 + c^2 \ge b^2.$

Both $a$ and $b$ are positive lengths, and the square function is increasing on the positive numbers, so $a^2 \ge b^2$ gives $a \ge b$. Equality holds exactly when the extra piece $c^2$ is $0$, that is, when $c = 0$, which is precisely the circle case.

The same equation, read the other way, gives $a^2 - c^2 = b^2$. Whenever $b > 0$, the right side is strictly positive, so $a^2 > c^2$, and since both are positive, $a > c$. That is why every genuine ellipse ($b > 0$, two distinct axis lengths) has $a$ strictly ahead of both $b$ and $c$, exactly the pattern both parts above happened to show.

Group the $x$ terms and the $y$ terms, and move the constant to the right:

$4x^2 - 16x + 25y^2 + 150y = -141.$

Factor the leading coefficient out of each group, since completing the square needs a bare $x^2$ and a bare $y^2$:

$4\left(x^2 - 4x\right) + 25\left(y^2 + 6y\right) = -141.$

Half of $-4$ is $-2$, and $(-2)^2 = 4$; half of $6$ is $3$, and $3^2 = 9$. Adding $4$ inside the first parentheses really adds $4 \cdot 4 = 16$ to the left side, and adding $9$ inside the second really adds $25 \cdot 9 = 225$, so add those same amounts to the right:

$4\left(x^2 - 4x + 4\right) + 25\left(y^2 + 6y + 9\right) = -141 + 16 + 225 = 100.$

Write each group as a square:

$4(x - 2)^2 + 25(y + 3)^2 = 100.$

Divide every term by $100$:

$\frac{(x - 2)^2}{25} + \frac{(y + 3)^2}{4} = 1.$

The center is $(2, -3)$, since $x - 2$ means $x - (2)$ and $y + 3$ means $y - (-3)$.

The larger denominator, $25$, sits under the $x$ term, so the major axis is horizontal:

$a^2 = 25, \qquad b^2 = 4 \quad\Rightarrow\quad a = 5, \; b = 2.$

Subtract for $c$:

$c^2 = a^2 - b^2 = 25 - 4 = 21 \quad\Rightarrow\quad c = \sqrt{21}.$

Measuring from the center $(2, -3)$ along the horizontal major axis, the vertices are $2 \pm 5$ on the $x$-coordinate, and the foci are $2 \pm \sqrt{21}$: vertices $(7, -3)$ and $(-3, -3)$, foci $(2 + \sqrt{21}, -3)$ and $(2 - \sqrt{21}, -3)$.

Divide $c$ by $a$:

$e = \frac{c}{a} = \frac{\sqrt{21}}{5} \approx 0.917.$

Eccentricity always satisfies $0 \le e < 1$, with $e = 0$ the circle and $e$ near $1$ a nearly flat curve. A value of about $0.92$ sits close to that upper end, so the foci are close to the vertices relative to the size of the ellipse, and the curve is noticeably flattened along its major axis rather than nearly round.

The stake separation is $2c$, and the string length is the constant sum of the two focal distances, $2a$:

$2c = 40 \;\Rightarrow\; c = 20, \qquad 2a = 58 \;\Rightarrow\; a = 29.$

Subtract for $b$:

$b^2 = a^2 - c^2 = 29^2 - 20^2 = 841 - 400 = 441 \quad\Rightarrow\quad b = 21.$

With the major axis horizontal, $a^2$ sits under $x^2$:

$\frac{x^2}{841} + \frac{y^2}{441} = 1.$

The lesson's derivation showed that the distance from a point of the ellipse to a focus runs between $a - c$ and $a + c$, with both extremes occurring at the vertices. Using $a = 29$ and $c = 20$:

$a - c = 29 - 20 = 9, \qquad a + c = 29 + 20 = 49.$

So the closest the bed's edge ever gets to a stake is $9$ feet, and the farthest is $49$ feet.

The string length is fixed at $58$ feet, and that alone fixes $a = 29$; nothing about moving the stakes changes the string. What moving the stakes changes is $2c$, the distance between them.

Eccentricity is

$e = \frac{c}{a}, \qquad a = 29 \text{ fixed}.$

With $a$ held fixed, shrinking $c$ shrinks $e$ toward $0$, and $e = 0$ is exactly the circle, the case where the two foci merge. So bringing the stakes closer together, which shrinks $c$, makes the bed rounder; spreading them apart would do the opposite, pushing $e$ toward $1$ and flattening the bed.

The lesson's derivation showed that the distance from a curve point to a focus runs from $a - c$ (nearest) to $a + c$ (farthest). Here the sun is the focus, so

$a - c = 3, \qquad a + c = 13.$

Add the two equations to eliminate $c$:

$2a = 16 \quad\Rightarrow\quad a = 8.$

Subtract them to eliminate $a$:

$2c = 10 \quad\Rightarrow\quad c = 5.$

Subtract for $b$:

$b^2 = a^2 - c^2 = 8^2 - 5^2 = 64 - 25 = 39 \quad\Rightarrow\quad b = \sqrt{39}.$

With the major axis horizontal and the center at the origin, $a^2 = 64$ sits under $x^2$:

$\frac{x^2}{64} + \frac{y^2}{39} = 1.$

A focus, where the sun sits, is $c$ units from the center along the major axis: $(5, 0)$ or, equally validly, $(-5, 0)$.

Divide:

$e = \frac{c}{a} = \frac{5}{8} = 0.625.$

Earth's $e \approx 0.017$ sits almost at the circular end of the scale, and Halley's $e \approx 0.97$ sits almost at the flattened end. A value of $0.625$ is well past the middle, much closer to Halley's end than to Earth's, so this comet's orbit is clearly elongated, an oval rather than a near-circle, though it still falls short of Halley's extreme shape.

The stake separation is $2c = 30$, so $c = 15$. The string length is $2a = 24$, so $a = 12$.

A genuine ellipse requires $a > c$, since the triangle inequality forces the sum of the two focal distances to be at least the distance between the foci, $2a \ge 2c$, with equality only when the curve collapses to the segment joining the stakes. Compare the two:

$a = 12 \;<\; c = 15.$

So even the collapsed case is out of reach: the string cannot stretch from one stake to the other at all, let alone loop around a peg with slack. No point satisfies the distance condition, so this plan produces no curve.

The stake spacing is unchanged at $30$ feet, so $c$ is still $15$. The new string gives $2a = 50$, so $a = 25$. Subtract for $b$:

$b^2 = a^2 - c^2 = 625 - 225 = 400 \quad\Rightarrow\quad b = 20.$

With the major axis horizontal, $a^2$ sits under $x^2$:

$\frac{x^2}{625} + \frac{y^2}{400} = 1.$

The relationship linking $a$, $b$, $c$ for an ellipse is $c^2 = a^2 - b^2$ (equivalently $a^2 = b^2 + c^2$), because $a$ is always the hypotenuse of the right triangle with legs $b$ and $c$, never a leg itself. Adding $a^2$ and $b^2$ has no place in this relationship at all.

Redo it correctly:

$c^2 = a^2 - b^2 = 625 - 400 = 225 \quad\Rightarrow\quad c = 15.$

So the foci are at $(\pm 15, 0)$. That also matches the original setup directly: the stakes were $30$ feet apart, half of that is $15$, exactly the value $c$ has to take, since the stake locations are the foci by construction. The trainee's $\sqrt{1025} \approx 32$ is not just arithmetically wrong, it places the foci outside the ellipse's own vertices at $x = \pm 25$, which is impossible for a focus.