Hyperbolas: Practice

Eccentricity is $e = \frac{c}{a}$, so $c = ea$.

$c = \frac{5}{4}(8) = 10$

Now use the hyperbola relation $b^2 = c^2 - a^2$.

$b^2 = 10^2 - 8^2 = 100 - 64 = 36$

Adding instead of subtracting gives $164$, and $100$ is $c^2$, not $b^2$.

Use the exact gap identity $\text{gap}(x) = \dfrac{ab}{x + \sqrt{x^2 - a^2}}$, with $a = 4$ and $b = 3$. First $\sqrt{25 - 16} = 3$.

$\text{gap}(5) = \frac{4 \cdot 3}{5 + 3} = \frac{12}{8} = \frac{3}{2}$

Check it directly: the line is at $\frac{3}{4}(5) = \frac{15}{4}$ and the curve is at $\frac{3}{4}\sqrt{25 - 16} = \frac{9}{4}$, and $\frac{15}{4} - \frac{9}{4} = \frac{6}{4} = \frac{3}{2}$. The gap is never $0$.

Group the variables and factor out the leading coefficients before completing the square.

$\bigl(x^2 + 6x\bigr) - 4\bigl(y^2 + 2y\bigr) = -1$

Half of $6$ is $3$, and half of $2$ is $1$, so the completed squares are $(x + 3)^2$ and $(y + 1)^2$:

$(x + 3)^2 - 4(y + 1)^2 = -1 + 9 - 4 = 4$

The centre is where both brackets vanish, at $(-3, -1)$. (Dividing by $4$ gives $\frac{(x+3)^2}{4} - (y+1)^2 = 1$.)

The positive term is the $y$ term, so $a^2 = 9$ and $b^2 = 4$, giving $a = 3$.

$c^2 = a^2 + b^2 = 9 + 4 = 13 \quad\Longrightarrow\quad c = \sqrt{13}$

Then

$e = \frac{c}{a} = \frac{\sqrt{13}}{3} \approx 1.20 ,$

which is greater than $1$, as it must be. Dividing by $b$ instead of $a$ gives $\frac{\sqrt{13}}{2}$, a common slip.

A hyperbola is rectangular exactly when $a = b$, which makes the asymptotes perpendicular. Divide $x^2 - y^2 = 16$ by $16$:

$\frac{x^2}{16} - \frac{y^2}{16} = 1 \quad\Longrightarrow\quad a = b = 4$

Its asymptotes are $y = \pm x$, whose slopes multiply to $-1$. Both $\frac{x^2}{9} - \frac{y^2}{4} = 1$ and $\frac{y^2}{4} - \frac{x^2}{9} = 1$ have $a \ne b$, and $\frac{x^2}{16} + \frac{y^2}{16} = 1$ is a circle, since both of its signs are positive.

Rectangular means $a = b$, so the relation $c^2 = a^2 + b^2$ becomes

$c^2 = a^2 + a^2 = 2a^2 \quad\Longrightarrow\quad c = a\sqrt{2} .$

Therefore

$e = \frac{c}{a} = \sqrt{2} \approx 1.414 ,$

whatever the size of the hyperbola. The $a$ cancels, so every rectangular hyperbola has the same eccentricity.

The foci are on the $x$-axis, so the form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $c = 13$ and slope $\frac{b}{a} = \frac{12}{5}$. Write $b = \frac{12}{5}a$ and substitute into $c^2 = a^2 + b^2$.

$169 = a^2 + \frac{144}{25}a^2 = \frac{169}{25}a^2 \quad\Longrightarrow\quad a^2 = 25$

So $a = 5$, $b = 12$, $b^2 = 144$, and the equation is $\frac{x^2}{25} - \frac{y^2}{144} = 1$. Check: $c^2 = 25 + 144 = 169$, so $c = 13$.

Substitute $x = 4$ and solve for $k$.

$\frac{16}{4} - \frac{k^2}{9} = 1 \quad\Longrightarrow\quad 4 - \frac{k^2}{9} = 1 \quad\Longrightarrow\quad \frac{k^2}{9} = 3$

So $k^2 = 27$ and, taking the positive root,

$k = \sqrt{27} = 3\sqrt{3} \approx 5.196 .$

Check: $\frac{16}{4} - \frac{27}{9} = 4 - 3 = 1$.

Rearrange the equation so that the $x$ term stands alone.

$\frac{x^2}{25} = 1 + \frac{y^2}{9} \ \ge\ 1$

A square is never negative, so the right side is at least $1$, forcing $x^2 \ge 25$ and therefore $|x| \ge 5$. The curve has a point at every $x$ with $|x| \ge 5$ and no point when $-5 < x < 5$. That empty vertical strip between the two vertices is exactly what splits the curve into two branches.

The foci are at $(\pm c, 0)$, so they are $2c$ apart. Find $c$ first.

$c^2 = a^2 + b^2 = 36 + 64 = 100 \quad\Longrightarrow\quad c = 10$

The distance between them is

$2c = 20 .$

The value $10$ is $c$ alone, and $12 = 2a$ is the transverse axis, not the focal separation.

The point $(4, 0)$ is on the $x$-axis, so it must be a vertex, which means the curve opens left and right with $a = 4$. For that form the slopes are $\pm\frac{b}{a}$.

$\frac{b}{4} = \frac{1}{2} \quad\Longrightarrow\quad b = 2 \quad\Longrightarrow\quad b^2 = 4$

The equation is $\frac{x^2}{16} - \frac{y^2}{4} = 1$, and substituting $(4, 0)$ gives $1 - 0 = 1$, so the point checks out.

Here $a = 3$ and $b = 4$, so $ab = 12$ and the bound reads $\text{gap}(x) \le \frac{12}{x}$. Force that bound below the target.

$\frac{12}{x} < 0.01 \quad\Longrightarrow\quad x > \frac{12}{0.01} = 1200$

Any $x > 1200$ works. Check that $x > 120$ is not enough: at $x = 120$ the gap is $\frac{12}{120 + \sqrt{14391}} \approx 0.050$, five times the target. No limit notation is needed anywhere: name a target, and the inequality hands back an $x$ that beats it.