Hyperbolas: Additional Practice (Set 1)

The foci are on the $y$-axis, so the curve opens up and down and $c = 10$. The constant difference is $2a$ by definition.

$2a = 12 \quad\Longrightarrow\quad a = 6$

Since $a = 6 < 10 = c$, the curve exists. Now find $b^2$ by subtracting:

$b^2 = c^2 - a^2 = 100 - 36 = 64$

The equation is $\frac{y^2}{36} - \frac{x^2}{64} = 1$. Note that $a^2 = 36$ goes under the positive term, and $c^2 = 100$ never appears in the equation at all.

The vertex is on the $y$-axis, so the curve opens up and down with $a = 5$. Rectangular means $a = b$, so $b = 5$ too.

$\frac{y^2}{25} - \frac{x^2}{25} = 1 \quad\Longrightarrow\quad y^2 - x^2 = 25$

Writing $x^2 - y^2 = 25$ instead would put the vertices on the $x$-axis, which contradicts the given vertex. The asymptotes are $y = \pm x$ either way, which is exactly why the asymptotes cannot tell the two apart.

Here $a = 6$ and $b = 8$, so $ab = 48$ and the bound is $\text{gap}(x) \le \frac{48}{x}$. Force the bound down to the target.

$\frac{48}{x} \le 0.01 \quad\Longrightarrow\quad x \ge \frac{48}{0.01} = 4800$

Check that $x \ge 480$ is not enough: at $x = 480$ the gap is about $\frac{48}{480 + 479.96} \approx 0.05$, five times too big. The gap never actually reaches $0$, but this inequality lets you beat any target you are given.

In standard form the equation is $\frac{y^2}{4} - \frac{x^2}{25} = 1$, so $a^2 = 4$ and $b^2 = 25$, and the positive term is the $y$ term.

$c^2 = a^2 + b^2 = 4 + 25 = 29 \quad\Longrightarrow\quad c = \sqrt{29}$

The foci lie on the $y$-axis, at $(0, \pm\sqrt{29}\,) \approx (0, \pm 5.39)$. Subtracting would give $\sqrt{21}$, which is the ellipse's rule, and putting them on the $x$-axis ignores which term is positive.

Group and complete both squares. Adding $9$ inside the first bracket adds $16(9) = 144$ to the left, and adding $4$ inside the second adds $(-1)(4) = -4$.

$16\bigl(x^2 + 6x + 9\bigr) - \bigl(y^2 - 4y + 4\bigr) = -124 + 144 - 4 = 16$

So $16(x + 3)^2 - (y - 2)^2 = 16$, and dividing by $16$ gives

$\frac{(x + 3)^2}{1} - \frac{(y - 2)^2}{16} = 1 .$

The centre is $(-3, 2)$ with $a = 1$ (from the positive term) and $b = 4$. The branches open left and right, so the vertices are $(-3 \pm 1,\ 2)$: the points $(-2, 2)$ and $(-4, 2)$. Here $a < b$, which is fine.

Completing the square puts the equation in the form $\frac{(x + 3)^2}{1} - \frac{(y - 2)^2}{16} = 1$, with centre $(-3, 2)$, $a = 1$, and $b = 4$. Replace the $1$ with $0$ and solve, keeping the brackets.

$\frac{(x + 3)^2}{1} = \frac{(y - 2)^2}{16} \quad\Longrightarrow\quad y - 2 = \pm 4(x + 3)$

The slopes are $\pm\frac{b}{a} = \pm\frac{4}{1} = \pm 4$, steep lines, and they pass through the centre $(-3, 2)$ rather than the origin.

Set the right side of each to $0$ and the two equations become identical, so the asymptote lines are the same:

$\frac{x^2}{9} = \frac{y^2}{16} \quad\Longrightarrow\quad y = \pm\frac{4}{3}x \ \text{ for both.}$

Everything else differs. The first has vertices $(\pm 3, 0)$, foci $(\pm 5, 0)$, and $e = \frac{5}{3}$; the second has vertices $(0, \pm 4)$, foci $(0, \pm 5)$, and $e = \frac{5}{4}$. They share the central rectangle, and therefore its diagonals, but they open along perpendicular axes.

Substitute $x = 6$ and solve for $y$.

$\frac{36}{4} - \frac{y^2}{9} = 1 \quad\Longrightarrow\quad 9 - \frac{y^2}{9} = 1 \quad\Longrightarrow\quad \frac{y^2}{9} = 8$

So $y^2 = 72$ and

$y = \pm\sqrt{72} = \pm 6\sqrt{2} \approx \pm 8.49 .$

Since $6 > a = 2$, the line does cut the right branch, and it cuts it twice, once above the $x$-axis and once below.

Put $x = a$ into the gap identity. The radical vanishes, since $\sqrt{a^2 - a^2} = 0$.

$\text{gap}(a) = \frac{ab}{a + \sqrt{a^2 - a^2}} = \frac{ab}{a} = b$

Read it straight off the picture instead and you get the same thing: the curve is at height $0$ at the vertex, while the line $y = \frac{b}{a}x$ is at height $\frac{b}{a} \cdot a = b$. That height $b$ is exactly the corner of the central rectangle, which is where the asymptote crosses the vertical line through the vertex. The gap is $b$ at the vertex and shrinks from there, but it never reaches $0$.

Dividing $b^2 = c^2 - a^2$ by $a^2$ shows that the asymptote slope depends on eccentricity alone.

$\frac{b}{a} = \sqrt{e^2 - 1}$

For $e = 1.05$ the slope is $\sqrt{0.1025} \approx 0.32$, so the asymptotes are nearly flat and the branches are narrow and sharply pointed. For $e = 3$ the slope is $\sqrt{8} \approx 2.83$, so the asymptotes are steep and the branches flare out. The value of $a$ only scales the picture; it never changes the slope, so eccentricity settles the question on its own.

The vertices give $a = 5$, so the equation is $\frac{x^2}{25} - \frac{y^2}{b^2} = 1$. Substitute the point $(13, 12)$.

$\frac{169}{25} - \frac{144}{b^2} = 1 \quad\Longrightarrow\quad \frac{144}{b^2} = \frac{169}{25} - 1 = \frac{144}{25}$

The numerators match, so $b^2 = 25$. This hyperbola is $x^2 - y^2 = 25$, a rectangular one, since $a = b = 5$: its asymptotes $y = \pm x$ are perpendicular and its eccentricity is $\sqrt{2}$.

The focus is on the $x$-axis, so the form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $c = 10$ and slope $\frac{b}{a} = \frac{4}{3}$, that is $b = \frac{4}{3}a$. Substitute into $c^2 = a^2 + b^2$.

$100 = a^2 + \frac{16}{9}a^2 = \frac{25}{9}a^2 \quad\Longrightarrow\quad a^2 = 36$

So $a = 6$, $b = 8$, and $b^2 = 64$, giving $\frac{x^2}{36} - \frac{y^2}{64} = 1$. The curve $\frac{x^2}{9} - \frac{y^2}{16} = 1$ has the right asymptote slope too, but its $c$ is $5$, so its focus sits at $(5, 0)$, not $(10, 0)$: the slope fixes only the ratio of $a$ to $b$, and the focus fixes the scale.