Hyperbolas: Free Response

The constant absolute difference is defined to be $2a$, so read it straight off the problem:

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

The foci are $(\pm 17, 0)$, and the distance from the center to a focus is $c$, so $c = 17$.

A hyperbola exists only when $0 < a < c$, since that inequality is forced by the strict triangle inequality applied to a point off the axis and the two foci. Here $a = 8$ and $c = 17$, and indeed $8 < 17$, so the constants are legal and the curve is genuine, not one of the degenerate cases.

The hyperbola relation, not the ellipse's, connects the three lengths:

$b^2 = c^2 - a^2 = 17^2 - 8^2 = 289 - 64 = 225.$

With $a^2 = 64$ and $b^2 = 225$, and the foci on the $x$-axis, the standard form is

$\frac{x^2}{64} - \frac{y^2}{225} = 1.$

The branches open left and right because the $x^2$ term is the one being added, not subtracted: it is the positive term, and $a^2$ always sits under the positive term regardless of which of $a^2$ and $b^2$ happens to be larger. Here $b^2 = 225$ is in fact larger than $a^2 = 64$, and that changes nothing about the orientation.

The vertices are $(\pm a, 0) = (\pm 8, 0)$, read directly from the standard form.

The asymptotes come from replacing the $1$ on the right of the standard-form equation with $0$ and solving for $y$, which is equivalent to the shortcut slope $\pm\frac{b}{a}$ for this horizontal orientation:

$y = \pm\frac{b}{a}x = \pm\frac{15}{8}x.$

The eccentricity is $e = \frac{c}{a} = \frac{17}{8}$, which is $2.125$.

A hyperbola's eccentricity measures how wide the branches flare: a value just above $1$ gives asymptotes close to the transverse axis and branches that are narrow and sharply pointed, while a larger value gives asymptotes far from that axis and branches that open widely. Since $\frac{17}{8}$ is more than twice $1$, this hyperbola's branches are on the wide, flared side, matching the fact that $b = 15$ is nearly twice $a = 8$.

Group and factor out the leading coefficients first:

$49(x^2 - 2x) - 36(y^2 + 4y) = 1859.$

Complete each square inside its own bracket. Half of $-2$ is $-1$, and $(-1)^2 = 1$, so add $1$ inside the first bracket; half of $4$ is $2$, and $2^2 = 4$, so add $4$ inside the second.

Now track what each addition does to the LEFT side, and match it on the right. Adding $1$ inside the first bracket, which is multiplied by $49$, adds $49$ to the left side. Adding $4$ inside the second bracket, which is multiplied by $-36$, adds $-36 \cdot 4 = -144$ to the left side, not $+144$:

$49(x^2 - 2x + 1) - 36(y^2 + 4y + 4) = 1859 + 49 - 144 = 1764.$

Write the completed squares and divide by $1764$:

$49(x - 1)^2 - 36(y + 2)^2 = 1764 \quad\Longrightarrow\quad \frac{(x - 1)^2}{36} - \frac{(y + 2)^2}{49} = 1.$

The center makes each squared bracket zero, so from $(x - 1)^2$ and $(y + 2)^2$ the center is $(1, -2)$.

The positive term is the $x$ term, so $a^2 = 36$ and $b^2 = 49$, giving $a = 6$ and $b = 7$. Then

$c^2 = a^2 + b^2 = 36 + 49 = 85 \quad\Longrightarrow\quad c = \sqrt{85}.$

The vertices sit $a$ units from the center along the transverse (horizontal) axis:

$(1 \pm 6, -2) = (7, -2) \ \text{and} \ (-5, -2).$

The foci sit $c$ units from the center along the same axis:

$(1 \pm \sqrt{85}, -2).$

For a shifted hyperbola the asymptote slope is the same $\pm\frac{b}{a}$ as the unshifted case, but the lines pass through the new center instead of the origin:

$y - (-2) = \pm\frac{7}{6}(x - 1) \quad\Longrightarrow\quad y + 2 = \pm\frac{7}{6}(x - 1).$

The eccentricity is

$e = \frac{c}{a} = \frac{\sqrt{85}}{6}.$

An irrational $c$ is not a red flag. The relation $c^2 = a^2 + b^2$ only guarantees that $c^2$ is a whole number when $a^2$ and $b^2$ are; it guarantees nothing about $c^2$ being a PERFECT square. Here $36 + 49 = 85$, and $85$ is not a perfect square, so $c = \sqrt{85}$ is the honest exact value, not a rounded approximation waiting to be cleaned up.

The stations are the foci, $60$ miles apart, so $2c = 60$ and $c = 30$. The constant distance difference is exactly $2a$, so $2a = 20$ and $a = 10$. Check legality: $10 < 30$, so the curve exists.

Then

$b^2 = c^2 - a^2 = 30^2 - 10^2 = 900 - 100 = 800.$

With the foci on the $x$-axis, the equation is

$\frac{x^2}{100} - \frac{y^2}{800} = 1.$

The points on the curve that lie on the line through the two foci are the vertices, $(\pm a, 0) = (\pm 10, 0)$.

Station A is at $(-30, 0)$ and station B is at $(30, 0)$. Check the vertex nearer A, at $(-10, 0)$, by the plain distance formula:

$\text{to A: } |-10 - (-30)| = 20, \qquad \text{to B: } |-10 - 30| = 40.$

The difference is $40 - 20 = 20$ miles, and the point is closer to A, matching the measurement without ever appealing to the value $2a$ that produced it.

Part B showed the vertex $(-10, 0)$, on the left branch, is $20$ miles closer to station A. By symmetry, every point $P$ of the left branch satisfies the same signed condition that produced the curve in part A:

$PF_2 - PF_1 = 20,$

where $F_1$ is station A and $F_2$ is station B, so the left branch, $x \le -10$, is the one consistent with "closer to A."

This single measurement can pin the event to that entire branch and no further, because a hyperbola is by definition the set of ALL points sharing one fixed difference of distances: the branch contains infinitely many locations, all consistent with the one number the timing difference supplied. To narrow the location further, a second, independent timing difference from a different pair of stations would produce a second hyperbola, and the event's true location must be one of the (typically one or two) points where the two curves intersect. One measurement gives a curve; a second narrows it to a small set of candidate points, which real direction-finding systems resolve using outside information such as a known rough region.

From the claim, $a = 22$ (the vertex distance) and $c = 21$ (the focus distance), so $a > c$. But a hyperbola requires $0 < a < c$, and here that fails, if only by a single unit: $22$ is not less than $21$.

The reason the inequality has to hold traces back to a genuine triangle. Take any point $P$ on the curve that does not lie on the line through the foci, so $P$, $F_1$, $F_2$ form a real triangle. The triangle inequality is strict for a real triangle, so each side is less than the sum of the other two:

$PF_1 < PF_2 + F_1F_2 \quad\text{and}\quad PF_2 < PF_1 + F_1F_2.$

Rearranging both gives $PF_1 - PF_2 < F_1F_2$ and $PF_2 - PF_1 < F_1F_2$, which together say $|PF_1 - PF_2| < F_1F_2$, that is $2a < 2c$. The claimed numbers violate exactly this, so no such hyperbola, and in fact no point off the axis satisfying the stated condition, can exist.

The transverse axis has length $2a$, so $2a = 40$ gives $a = 20$. Since $20 < 21 = c$, this is legal.

$b^2 = c^2 - a^2 = 21^2 - 20^2 = 441 - 400 = 41.$

The standard-form equation is

$\frac{x^2}{400} - \frac{y^2}{41} = 1.$

Note how close this legal hyperbola sits to the boundary: $a = 20$ and the illegal claim's $a = 22$ sit one unit on opposite sides of $c = 21$, and that single unit is the entire difference between a curve that exists and one that does not.

In general, the inequality of part A is $0 < a < c$ for every hyperbola, no matter its size. Dividing all sides by the positive number $a$ preserves the direction:

$\frac{a}{a} < \frac{c}{a} \quad\Longrightarrow\quad 1 < \frac{c}{a} = e.$

So $e > 1$ always, and this is not a coincidence about any one hyperbola: it is a direct consequence of the same triangle inequality that ruled out the claim in part A.

An ellipse is governed by the opposite condition. Its distance condition is a SUM, not a difference, and the argument that bounds it produces $a > c$ instead of $a < c$ (the sum of the two focal distances is bounded BELOW by the distance between the foci, the reverse relationship from a difference, which is bounded above by it). Dividing $a > c$ by the positive number $a$ gives $1 > \frac{c}{a}$, so an ellipse's eccentricity is always below $1$. The two curves' constants obey opposite inequalities, and eccentricity inherits that opposition directly.

Read $a^2 = 121$ and $b^2 = 3600$ from the denominators, so $a = 11$ and $b = 60$. Then

$c^2 = a^2 + b^2 = 121 + 3600 = 3721 = 61^2 \quad\Longrightarrow\quad c = 61.$

The eccentricity is $e = \frac{c}{a} = \frac{61}{11}$, and since the $x^2$ term is positive, the asymptote slopes are $\pm\frac{b}{a} = \pm\frac{60}{11}$.

The new hyperbola has $a_{\text{new}} = ka$ and $b_{\text{new}} = kb$, and it obeys the same relation as every hyperbola:

$c_{\text{new}}^2 = a_{\text{new}}^2 + b_{\text{new}}^2 = (ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2 c^2.$

Since $k > 0$, taking the positive square root gives $c_{\text{new}} = kc$, not merely $c_{\text{new}}^2 = k^2c^2$ with an ambiguous sign; both $c_{\text{new}}$ and $kc$ are positive, so they are equal outright.

Now form the new eccentricity, and the factor of $k$ cancels because it multiplies both the numerator and the denominator:

$e_{\text{new}} = \frac{c_{\text{new}}}{a_{\text{new}}} = \frac{kc}{ka} = \frac{c}{a} = e.$

This holds for every positive $k$, so scaling $a$ and $b$ by the same factor never changes the eccentricity, whatever that factor is.

Eccentricity is built only from $a$ and $b$, the shape of the hyperbola, and never from its center or orientation. When two rectangular hyperbolas happen to share a center and orientation, part B applies directly: the second is obtained from the first by scaling with $k = a'/a$, because $ka = a'$ and $kb = ka = a' = b'$, so they share the same eccentricity. But part B's argument alone does not cover two rectangular hyperbolas with different centers or orientations, so the general claim needs the direct computation below, which depends on $a$ and $b$ alone and is therefore true for every rectangular hyperbola regardless of where it sits or how it is oriented.

That shared value is found directly, without needing a second example: for $a = b$,

$c^2 = a^2 + b^2 = 2a^2 \quad\Longrightarrow\quad c = a\sqrt{2} \quad\Longrightarrow\quad e = \frac{c}{a} = \sqrt{2},$

and $a$ has canceled out of the ratio entirely, so this value does not depend on which rectangular hyperbola was chosen.

What scaling by $k$ DOES change is everything measured in actual length: $a$ becomes $ka$, $b$ becomes $kb$, $c$ becomes $kc$ by part B, so the distance between the foci ($2c$) and between the vertices ($2a$) both scale by $k$ as well. The curve after scaling is geometrically similar to the original, same angles, same proportions, but a different size, and only when $k = 1$ is it the identical curve. Eccentricity is a SHAPE quantity, and shape survives the scaling that size does not.