Hyperbolas: Practice

The orientation is decided by which term is positive, and here that is the $x^2$ term. Setting $y = 0$ finds the turning points.

$\frac{x^2}{16} = 1 \quad\Longrightarrow\quad x = \pm 4$

The curve turns back at $(\pm 4, 0)$ on the $x$-axis, so the branches open left and right. Setting $x = 0$ gives $-\frac{y^2}{9} = 1$, which has no real solution, confirming the curve never meets the $y$-axis.

The $x^2$ term is positive, so the vertices lie on the $x$-axis and $a^2 = 25$.

$a = 5 \quad\Longrightarrow\quad \text{vertices } (\pm 5, 0)$

The number $12$ is $b$, which never touches the curve, and $13$ is $c$, which locates the foci, not the vertices.

The positive term is the $x^2$ term, so the foci lie on the $x$-axis at $(\pm c, 0)$. Find $c$ by adding.

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

The foci are $(\pm 5, 0)$. They sit outside the vertices $(\pm 3, 0)$, exactly as $c > a$ requires.

Replace the $1$ on the right with $0$ and solve. That turns the equation into a difference of squares, which factors.

$\frac{x^2}{9} - \frac{y^2}{16} = 0 \quad\Longrightarrow\quad y^2 = \frac{16}{9}x^2 \quad\Longrightarrow\quad y = \pm\frac{4}{3}x$

The slopes are $\pm\frac{b}{a} = \pm\frac{4}{3}$, using the square roots $b = 4$ and $a = 3$, not the denominators $16$ and $9$.

The derivation defines $b$ by $b^2 = c^2 - a^2$, which is legal precisely because the distance condition forces $c > a$. Rearranging gives

$c^2 = a^2 + b^2 .$

So $c$ is the hypotenuse of a right triangle with legs $a$ and $b$, and $c$ is the largest of the three. (For an ellipse it is $a$ that is largest, with $a^2 = b^2 + c^2$.)

A hyperbola is the set of points $P$ satisfying

$\bigl|\, PF_1 - PF_2 \,\bigr| = 2a .$

A constant sum gives an ellipse instead. The absolute value is what creates two branches: $PF_1 - PF_2 = 2a$ describes the branch nearer $F_2$, and $PF_1 - PF_2 = -2a$ describes the branch nearer $F_1$.

For a hyperbola, $b^2 = c^2 - a^2$. The larger number, $c$, comes first.

$b^2 = 5^2 - 3^2 = 25 - 9 = 16$

Adding would give $34$, which is the ellipse-versus-hyperbola sign error. The value $4$ is $b$, not $b^2$, and $9$ is $a^2$.

The shifted form is $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ with centre $(h, k)$. Match the given equation to it by reading the value that makes each squared bracket zero.

$x - 2 = 0 \ \Rightarrow\ h = 2, \qquad y + 5 = y - (-5) = 0 \ \Rightarrow\ k = -5$

The centre is $(2, -5)$. The sign inside the bracket is always the opposite of the coordinate.

Substitute $x = 0$ and see what the equation demands of $y$.

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

No real number has a negative square, so there is no $y$-intercept. More strongly, $\frac{x^2}{9} = 1 + \frac{y^2}{16} \ge 1$ forces $|x| \ge 3$, so the whole strip $-3 < x < 3$ is empty. That empty strip is what separates the two branches.

The transverse axis is the segment joining the two vertices, and each vertex is $a$ from the centre. Here $a^2 = 25$, so $a = 5$.

$\text{length} = 2a = 2(5) = 10$

The vertices are $(\pm 5, 0)$, which are $10$ units apart. The value $6 = 2b$ is the conjugate axis, which touches no point of the curve.

The distance condition forces $c > a$ for every hyperbola, because the difference of two sides of a triangle cannot reach the third side. Dividing that inequality by the positive number $a$ gives

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

Eccentricity is what sorts the conics: an ellipse has $e < 1$, a parabola has $e = 1$, and a hyperbola has $e > 1$.

The constant absolute difference of the distances is $2a$ by definition, and here that constant is $8$.

$2a = 8 \quad\Longrightarrow\quad a = 4$

Check that the curve exists: the foci give $c = 5$, and $a = 4 < 5 = c$, so the condition $0 < a < c$ holds. (Had the difference been $12$, we would need $a = 6 > c$, and no point would qualify.)