Hyperbolas: Additional Practice (Set 2)

For a horizontal hyperbola the asymptote slope is $\frac{b}{a} = \sqrt{e^2 - 1}$, so square both sides.

$e^2 - 1 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \quad\Longrightarrow\quad e^2 = \frac{25}{16}$

Taking the positive root, $e = \frac{5}{4} = 1.25$, which is greater than $1$ as required. (Concretely: take $a = 4$ and $b = 3$, so $c = 5$ and $e = \frac{5}{4}$.)

The $y^2$ term is positive, so $a^2 = 16$ and $a = 4$ (not $3$, whatever the denominators look like).

$c^2 = a^2 + b^2 = 16 + 9 = 25 \quad\Longrightarrow\quad c = 5$

So

$e = \frac{c}{a} = \frac{5}{4} = 1.25 > 1 .$

Dividing by $b = 3$ instead of $a = 4$ gives $\frac{5}{3}$, the eccentricity of the other hyperbola with the same numbers, $\frac{x^2}{9} - \frac{y^2}{16} = 1$.

For the vertical form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the asymptote slope is $\frac{a}{b}$, not $\frac{b}{a}$. So $\frac{a}{b} = \frac{3}{4}$, and taking $a = 3$, $b = 4$ gives

$c^2 = a^2 + b^2 = 9 + 16 = 25 \quad\Longrightarrow\quad c = 5, \qquad e = \frac{c}{a} = \frac{5}{3} .$

Compare the previous question: the same asymptotes give $e = \frac{5}{4}$ horizontally but $e = \frac{5}{3}$ vertically. Asymptotes alone never determine a hyperbola, because they never reveal which term is positive.

Divide every term by $100$ so the right side becomes $1$.

$\frac{25y^2}{100} - \frac{4x^2}{100} = 1 \quad\Longrightarrow\quad \frac{y^2}{4} - \frac{x^2}{25} = 1$

The denominators come out inverted from the coefficients: the $25$ in front of $y^2$ becomes a $4$ underneath it. The $y^2$ term is positive, so this hyperbola opens up and down, with $a = 2$ and $b = 5$.

Substitute $x = 5$ into the equation and solve for $y$.

$\frac{25}{9} - \frac{y^2}{16} = 1 \quad\Longrightarrow\quad \frac{y^2}{16} = \frac{25}{9} - 1 = \frac{16}{9}$

So $y^2 = \frac{256}{9}$ and $y = \pm\frac{16}{3}$. In general this half-height equals $\frac{b^2}{a} = \frac{16}{3}$, which is why the chord through a focus is a quick way to plot two more points when sketching.

The transverse axis has length $2a$, so $2a = 8$ and $a = 4$. Then $e = \frac{c}{a}$ gives

$c = ea = 3(4) = 12 .$

Now subtract, because $b^2 = c^2 - a^2$ for a hyperbola:

$b^2 = 144 - 16 = 128$

Adding instead would give $160$. As a check, the asymptote slope is $\frac{b}{a} = \sqrt{e^2 - 1} = \sqrt{8}$, and indeed $\frac{\sqrt{128}}{4} = \frac{8\sqrt{2}}{4} = 2\sqrt{2} = \sqrt{8}$.

Substitute $y = 3$ and solve for $x$.

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

So $x = \pm 2\sqrt{2} \approx \pm 2.83$. Both solutions are real, one on each branch, and both satisfy $|x| \ge a = 2$, as every point of the curve must. A horizontal line always meets a left-right hyperbola twice, one hit per branch.

Here $a = 4$, $b = 3$, and $c = \sqrt{16 + 9} = 5$, so $e = \frac{5}{4}$. The proof's converse gives the focal distances as $PF_1 = |ex + a|$ and $PF_2 = |ex - a|$, and on the right branch both are positive.

$PF_2 = \frac{5}{4}(8) - 4 = 10 - 4 = 6$

Check it with the distance formula: at $x = 8$, $y^2 = 9\left(\frac{64}{16} - 1\right) = 27$, and the distance to $(5, 0)$ is $\sqrt{(8-5)^2 + 27} = \sqrt{36} = 6$. The value $14$ is the distance to the left focus, and the two differ by $2a = 8$, as they must.

The definition says $|PF_1 - PF_2| = 6$, so removing the absolute value gives two cases.

$11 - PF_2 = 6 \ \Rightarrow\ PF_2 = 5, \qquad 11 - PF_2 = -6 \ \Rightarrow\ PF_2 = 17$

Both are genuine. With $a = 3$ and $c = 5$ the eccentricity is $e = \frac{5}{3}$, and the focal distances are $PF_1 = |ex + a|$ and $PF_2 = |ex - a|$. Taking $x = \frac{24}{5}$ (right branch) gives $PF_1 = 11$ and $PF_2 = 5$; taking $x = -\frac{42}{5}$ (left branch) gives $PF_1 = 11$ and $PF_2 = 17$. Dropping the absolute value would lose one of the two branches, and with it one of the two answers.

Test each point in the equation. The vertices $(\pm 3, 0)$ give $\frac{9}{9} - 0 = 1$, so both are on the curve, and $\left(5, \frac{16}{3}\right)$ gives

$\frac{25}{9} - \frac{256/9}{16} = \frac{25}{9} - \frac{16}{9} = 1 . \ \checkmark$

But $(0, 4)$ gives

$0 - \frac{16}{16} = -1 \ne 1 .$

No point of this curve has $x = 0$: the relation $\frac{x^2}{9} = 1 + \frac{y^2}{16} \ge 1$ forces $|x| \ge 3$. The point $(0, 4)$ is the tip of the conjugate axis, which lies in the empty strip between the branches.

Both the focus and the vertex lie on the $y$-axis, so the curve opens up and down, with $c = 7$ and $a = 4$. (The focus is further from the centre than the vertex, exactly as $c > a$ demands.)

$b^2 = c^2 - a^2 = 49 - 16 = 33$

So the equation is $\frac{y^2}{16} - \frac{x^2}{33} = 1$. The denominators of a hyperbola are $a^2$ and $b^2$; $c^2 = 49$ never appears in the equation itself.

With $0$ on the right, the left side is a difference of squares and factors.

$\left(\frac{x}{2} - \frac{y}{3}\right)\left(\frac{x}{2} + \frac{y}{3}\right) = 0 \quad\Longrightarrow\quad y = \pm\frac{3}{2}x$

A product is zero exactly when one of its factors is, so the graph is the pair of lines $y = \pm\frac{3}{2}x$, which are precisely the asymptotes of $\frac{x^2}{4} - \frac{y^2}{9} = 1$. That is why replacing the $1$ with $0$ is a legitimate way to find asymptotes rather than a memory trick.