Hyperbolas: Practice

Rectangular means $a = b$, so $c^2 = a^2 + b^2 = 2a^2$. The focus gives $c = 6$.

$2a^2 = 36 \quad\Longrightarrow\quad a^2 = 18 = b^2$

The equation is $\frac{x^2}{18} - \frac{y^2}{18} = 1$, that is $x^2 - y^2 = 18$. Check: $a = b = \sqrt{18}$, so $c = \sqrt{18 + 18} = 6$, and $e = \frac{6}{\sqrt{18}} = \sqrt{2}$, as every rectangular hyperbola must have.

Group and complete both squares. Adding $1$ inside the first bracket adds $4$ to the left, and adding $4$ inside the second adds $(-9)(4) = -36$.

$4\bigl(x^2 - 2x + 1\bigr) - 9\bigl(y^2 - 4y + 4\bigr) = 32 + 4 - 36 = 0$

The constant collapses to $0$, so there is no $1$ to divide by:

$4(x - 1)^2 = 9(y - 2)^2 \quad\Longrightarrow\quad y - 2 = \pm\frac{2}{3}(x - 1)$

That is a degenerate conic: two straight lines crossing at $(1, 2)$, which are exactly the asymptotes the hyperbola would have had. Whenever the constant lands on zero, the hyperbola collapses into its own asymptotes.

The vertices are on the $x$-axis, so $a = 2$ and the form is $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$. Substitute the point, remembering that $(3\sqrt{3})^2 = 9 \cdot 3 = 27$.

$\frac{16}{4} - \frac{27}{b^2} = 1 \quad\Longrightarrow\quad \frac{27}{b^2} = 3 \quad\Longrightarrow\quad b^2 = 9$

So the hyperbola is $\frac{x^2}{4} - \frac{y^2}{9} = 1$. Reading $b^2 = 27$ straight off the squared $y$-coordinate instead produces $\frac{x^2}{4} - \frac{y^2}{27} = 1$, which fails the check: it sends the point to $4 - 1 = 3$, not $1$.

The defining property is $|PF_1 - PF_2| = 2a = 6$. On the right branch the point is closer to the right focus, so $PF_1 - PF_2 = +6$.

$10 - PF_2 = 6 \quad\Longrightarrow\quad PF_2 = 4$

The distance is $4$, which is legal because it exceeds the minimum $c - a = 2$. Adding instead of subtracting gives $16$, the answer for a point on the left branch, where the sign of the difference flips.

Take any point $P$ of the curve that is off the line through the foci. It forms a genuine, non-degenerate triangle with $F_1$ and $F_2$, and in a genuine triangle the triangle inequality is strict, so the difference of two sides is strictly less than the third.

$|PF_1 - PF_2| < F_1F_2 \quad\Longrightarrow\quad 2a < 2c$

A hyperbola therefore needs $0 < a < c$. The pair $a = 5$, $c = 4$ violates it: no point can be $10$ units further from one focus than from the other when the foci are only $8$ apart, so the set is empty. The other three all satisfy $a < c$, even $a = 3$ with $c = 3.01$, which just gives a very narrow, sharply pointed pair of branches.

Use the exact identity $\text{gap}(x) = \dfrac{ab}{x + \sqrt{x^2 - a^2}}$ with $a = 5$ and $b = 12$, so $ab = 60$. First $\sqrt{169 - 25} = \sqrt{144} = 12$.

$\text{gap}(13) = \frac{60}{13 + 12} = \frac{60}{25} = \frac{12}{5}$

Direct check: the line is at $\frac{12}{5}(13) = 31.2$ and the curve is at $\frac{12}{5}(12) = 28.8$, and the difference is $2.4 = \frac{12}{5}$. Notice $x = 13$ is the focus's $x$-coordinate here, and the gap is still comfortably positive.

The exact gap on the upper right is

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

and both facts read straight off it. The numerator $ab$ is a fixed positive number and the denominator is positive for every $x \ge a$, so the gap is never $0$: no touching and no crossing. But the denominator is at least $x$, so $\text{gap}(x) \le \frac{ab}{x}$; name any target $t > 0$ and every $x > \frac{ab}{t}$ makes the gap smaller than $t$. Approaching a line and reaching it are different claims, and only approaching is true here.

The asymptotes cross at the centre $(2, 3)$. The vertex $(2, 7)$ sits directly above the centre, so the hyperbola opens up and down, and

$a = 7 - 3 = 4 .$

For the vertical form the slope is $\frac{a}{b} = 2$, so $b = \frac{a}{2} = 2$ and $b^2 = 4$. The equation is

$\frac{(y - 3)^2}{16} - \frac{(x - 2)^2}{4} = 1 .$

Check the asymptotes: $y - 3 = \pm 2(x - 2)$ gives $y = 2x - 1$ and $y = -2x + 7$, as required.

Rectangular with $a = b$ means the equation is $x^2 - y^2 = a^2$. Substitute the point.

$5^2 - 3^2 = 25 - 9 = 16 = a^2 \quad\Longrightarrow\quad a = b = 4$

Then $c^2 = a^2 + b^2 = 16 + 16 = 32$, so $c = \sqrt{32} = 4\sqrt{2}$ and the foci are $(\pm 4\sqrt{2},\ 0) \approx (\pm 5.66,\ 0)$. As a check, $e = \frac{4\sqrt{2}}{4} = \sqrt{2}$, the eccentricity every rectangular hyperbola has.

Vertices on the $y$-axis means the $y^2$ term is positive, so only $\frac{y^2}{4} - \frac{x^2}{9} = 1$ and $\frac{y^2}{9} - \frac{x^2}{4} = 1$ are candidates. For the vertical form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ the slopes are $\pm\frac{a}{b}$.

$\frac{y^2}{4} - \frac{x^2}{9} = 1 \ \Rightarrow\ \frac{a}{b} = \frac{2}{3} \ \checkmark \qquad \frac{y^2}{9} - \frac{x^2}{4} = 1 \ \Rightarrow\ \frac{a}{b} = \frac{3}{2} \ \times$

So the answer is $\frac{y^2}{4} - \frac{x^2}{9} = 1$. The curve $\frac{x^2}{9} - \frac{y^2}{4} = 1$ does have slopes $\pm\frac{2}{3}$ as well, but its vertices sit on the $x$-axis, which is why asymptotes alone never pin down a hyperbola.

Group and complete the squares. Adding $4$ inside the $y$ bracket adds $4$; adding $9$ inside the $x$ bracket adds $(-4)(9) = -36$.

$\bigl(y^2 - 4y + 4\bigr) - 4\bigl(x^2 - 6x + 9\bigr) = 36 + 4 - 36 = 4$

So $(y - 2)^2 - 4(x - 3)^2 = 4$, and dividing by $4$ gives

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

The $y$ term is positive, so the curve opens up and down from the centre $(3, 2)$, with $a = 2$ and $b = 1$. Then $c^2 = 4 + 1 = 5$, and the foci are $(3,\ 2 \pm \sqrt{5}\,)$, moved vertically from the centre.

A hyperbola needs the two denominators to have opposite signs, so that one term is effectively positive and the other negative. That means their product is negative.

$(k - 2)(k - 6) < 0 \quad\Longrightarrow\quad 2 < k < 6$

Inside that interval $k - 2 > 0$ and $k - 6 < 0$, so the equation reads $\frac{x^2}{k-2} - \frac{y^2}{6-k} = 1$, a hyperbola opening left and right. If $k > 6$ both denominators are positive and you get an ellipse; if $k < 2$ both are negative and no point satisfies the equation; and $k = 2$ or $k = 6$ makes a denominator zero.