Lesson Objectives

- Learn how to find the foci of a hyperbola
- Learn how to find the vertices and covertices of a hyperbola
- Learn how to write the equation of a hyperbola
- Learn how to find the asymptotes and fundamental rectangle of a hyperbola
- Learn how to sketch the graph of a hyperbola

## Equations and Graphs of Hyperbolas

In this lesson, we will continue to learn about conic sections with a focus on the hyperbola. In the last lesson, we learned that the ellipse was defined as the set of all points in a plane the sum of whose distances from two fixed points, known as the foci, is constant. Similarly, the hyperbola is defined as the set of all points in a plane where the absolute value of the difference of the distances from two fixed points, known as the foci, is constant.

Note: The substitution we made above often causes confusion. Where does b

To find the y-intercepts, plug in a 0 for x: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$\frac{0^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$-\frac{y^2}{b^2} = 1$$ $$y^2 = -b^2$$ The above equation does not have any real number solutions, therefore, we will not have any y-intercepts in this case.

(a, b), (-a, b), (a, -b), and (-a, -b). The slopes of the diagonals of the fundamental rectangle are ±b/a. We can extend these diagonals to obtain the asymptotes that we found above of y = ±(b/a)x.

Example #1: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$\frac{x^2}{9} - \frac{y^2}{25} = 1$$ First, let's match this to our equation from above. We can see the center is at the origin and the transverse axis will lie on the x-axis. $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$\frac{x^2}{9} - \frac{y^2}{25} = 1$$ $$a^2 = 9, b^2 = 25$$ $$c^2 = a^2 + b^2$$ $$c^2 = 9 + 25 = 34$$ Note: a, b, and c are positive as defined by our derivation above. Take this into consideration when dealing with square roots in this section. $$a = \sqrt{9} = 3$$ $$\text{Vertices:} \, (3, 0), (-3, 0)$$ $$b = \sqrt{25} = 5$$ $$\text{Covertices:} \, (0, 5), (0, -5)$$ $$c = \sqrt{34}$$ $$\text{Foci:} \, (\sqrt{34}, 0), (-\sqrt{34}, 0)$$ $$\text{Asymptotes:} \, y = \pm \frac{5}{3}x$$ Desmos Link for More Detail Example #2: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$-9x^2 + 16y^2 - 144 = 0$$ Let's first write our equation in standard form. $$-9x^2 + 16y^2 - 144 = 0$$ Add 144 to both sides: $$-9x^2 + 16y^2 = 144$$ Divide each side by 144: $$\frac{-9}{144}x^2 + \frac{16}{144}y^2 = \frac{144}{144}$$ $$-\frac{x^2}{16} + \frac{y^2}{9} = 1$$ Rearrange: $$\frac{y^2}{9} -\frac{x^2}{16} = 1$$ We can see the center is at the origin and the transverse axis will lie on the y-axis. $$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ $$\frac{y^2}{9} -\frac{x^2}{16} = 1$$ $$a^2 = 9, b^2 = 16$$ $$c^2 = a^2 + b^2$$ $$c^2 = 9 + 16 = 25$$ $$a = \sqrt{9} = 3$$ $$\text{Vertices:} \, (0, 3), (0, -3)$$ $$b = \sqrt{16} = 4$$ $$\text{Covertices:} \, (4, 0), (-4, 0)$$ $$c = \sqrt{25} = 5$$ $$\text{Foci:} \, (0, 5), (0, -5)$$ $$\text{Asymptotes:} \, y = \pm \frac{3}{4}x$$ Desmos Link for More Detail

We can use the following table when considering translations with hyperbolas.

If h and k are positive real numbers (h > 0 and k > 0):

Let's look at an example.

Example #3: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{9} = 1$$ Let's think about matching our form from above. $$1) \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ $$2) \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ Here k = 1, and h = -1, so the center will occur at (-1, 1). In other words, since y has been replaced with (y - 1) and x has been replaced with (x + 1), the center has been shifted 1 unit left and 1 unit up and will now occur at (-1, 1).

The vertices, covertices, and foci are a, b, and c units away from the center, respectively. $$a^2 = 16, a = 4$$ $$b^2 = 9, b = 3$$ $$c^2 = a^2 + b^2 = 16 + 9 = 25, c = 5$$ Since the transverse axis is vertical, to find the vertices, we will start at the center and move up by a units and down by a units. The x-coordinate from the center will not change. $$(-1, 1 + 4) = (-1, 5)$$ $$(-1, 1 - 4) = (-1, -3)$$ $$\text{Vertices:} \, (-1, -3), (-1, 5)$$ Similarly, to find the foci, we will use the same thought process. We will start at the center and move up by c units and down by c units. The x-coordinate from the center will not change. $$(-1, 1 + 5) = (-1, 6)$$ $$(-1, 1 - 5) = (-1, -4)$$ $$\text{Foci:} \, (-1, 6), (-1, -4)$$ To find the covertices we will need to move horizontally since the conjugate axis is horizontal. We will start at the center and move left by b units and right by b units. The y-coordinate from the center will not change. $$(-1 + 3, 1) = (2, 1)$$ $$(-1 - 3, 1) = (-4, 1)$$ $$\text{Covertices:} \, (2, 1), (-4, 1)$$ To find the asymptotes, we will use the point-slope formula. $$y - y_1 = m(x - x_1)$$ Let's go back to our original form for a moment. $$1) \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ Here, we have asymptotes: y = ±(a/b)x. We know each asymptote passes through the origin, which is the center in this case. $$2) \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ Here, the slope of the asymptotes will remain ±(a/b) and each asymptote will pass through the center, which is (h, k). $$y - y_1 = m(x - x_1)$$ $$y - k = \pm \frac{a}{b}(x - h)$$ $$y = \pm \frac{a}{b}(x - h) + k$$ If we plug in for a, b, h, and k, we get: $$y = \pm \frac{4}{3}(x + 1) + 1$$ If we compare this to: $$y = \pm \frac{4}{3}x$$ We have a shift left by 1 unit coming from (x + 1) and a shift up by 1 unit coming from (+1). This matches the shift from a center of (0, 0) to our current center of (-1, 1).

Asymptote #1 with a slope of 4/3: $$y = \frac{4}{3}(x + 1) + 1$$ $$y = \frac{4}{3}x + \frac{4}{3} + \frac{3}{3}$$ $$y = \frac{4}{3}x + \frac{7}{3}$$ Asymptote #2 with a slope of -4/3: $$y = -\frac{4}{3}(x + 1) + 1$$ $$y = -\frac{4}{3}x - \frac{4}{3} + \frac{3}{3}$$ $$y = -\frac{4}{3}x - \frac{1}{3}$$ The endpoints of the fundamental rectangle can be found as: $$(h + b, k + a) = (-1 + 3, 1 + 4) = (2, 5)$$ $$(h + b, k - a) = (-1 + 3, 1 - 4) = (2, -3)$$ $$(h - b, k + a) = (-1 - 3, 1 + 4) = (-4, 5)$$ $$(h - b, k - a) = (-1 - 3, 1 - 4) = (-4, -3)$$ Desmos Link for More Detail

Additionally, as we saw in the example above the equations of the asymptotes can be found using point-slope formula. $$y - y_1 = m(x - x_1)$$ The center (h, k) is used as the point (x

Let's look at an example.

Example #4: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$4x^2 - 9y^2 + 8x - 18y - 41 = 0$$ Write in standard form. $$4x^2 + 8x - 9y^2 - 18y - 41 = 0$$ $$4(x^2 + 2x) - 9(y^2 + 2y) = 41$$ $$4(x^2 + 2x + 1 - 1) - 9(y^2 + 2y + 1 - 1) = 41$$ $$4(x^2 + 2x + 1) + (4)(-1) - 9(y^2 + 2y + 1) + (-9)(-1) = 41$$ $$4(x + 1)^2 -4 - 9(y + 1)^2 + 9 = 41$$ $$4(x + 1)^2 - 9(y + 1)^2 + 5 = 41$$ $$4(x + 1)^2 - 9(y + 1)^2 = 36$$ $$\frac{4(x + 1)^2}{36} - \frac{9(y + 1)^2}{36} = \frac{36}{36}$$ $$\frac{(x + 1)^2}{9} - \frac{(y + 1)^2}{4} = 1$$ Match the form and grab the formulas from the table: $$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$ $$h = -1, k = -1$$ $$\text{Center:} \, (-1, -1)$$ $$\text{Vertices:} \, (h - a, k), (h + a, k)$$ $$a^2 = 9, a = 3$$ $$b^2 = 4, b = 2$$ $$h - a = -1 - 3 = -4$$ $$h + a = -1 + 3 = 2$$ $$\text{Vertices:} \, (-4, -1), (2, -1)$$ $$\text{Covertices:} \, (h, k - b), (h, k + b)$$ $$k - b = -1 - 2 = -3$$ $$k + b = -1 + 2 = 1$$ $$\text{Covertices:} \, (-1, -3), (-1, 1)$$ $$\text{Foci:} \, (h - c, k), (h + c, k)$$ $$c^2 = 13, c = \sqrt{13}$$ $$h - c = -1 - \sqrt{13}$$ $$h + c = -1 + \sqrt{13}$$ $$\text{Foci:} \, (-1 - \sqrt{13}, -1), (-1 + \sqrt{13}, -1)$$ $$\text{Asymptotes:} \, y = \pm \frac{b}{a}(x - h) + k$$ $$y = \pm \frac{2}{3}(x + 1) - 1$$ $$y = \frac{2}{3}x + \frac{2}{3} - \frac{3}{3}$$ $$y = \frac{2}{3}x - \frac{1}{3}$$ $$y = -\frac{2}{3}x - \frac{2}{3} - \frac{3}{3}$$ $$y = -\frac{2}{3}x - \frac{5}{3}$$ Endpoints for the Fundamental Rectangle: $$(h + a, k + b) = (-1 + 3, -1 + 2) = (2, 1)$$ $$(h - a, k + b) = (-1 - 3, -1 + 2) = (-4, 1)$$ $$(h + a, k - b) = (-1 + 3, -1 - 2) = (2, -3)$$ $$(h - a, k - b) = (-1 - 3, -1 - 2) = (-4, -3)$$ Desmos Link for More Detail

Example #5: Find the standard form of the equation of each hyperbola. $$\text{Vertices:} \, (6, 4), (6, -16)$$ $$\text{Foci:} \, (6, -6 + 2\sqrt{34}), (6, -6 - 2\sqrt{34})$$ Let's start by finding the center. To do this, we will use the midpoint formula with the vertices. $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ $$M = \left(\frac{6 + 6}{2}, \frac{4 + (-16)}{2}\right) = (6, -6)$$ $$\text{Center:} \, (6, -6)$$ Since the y-coordinates of the vertices and foci are different while the x-coordinates remain the same, we know that the transverse axis is vertical. $$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ Plug in for h and k. Here h = 6 and k = -6. $$\frac{(y + 6)^2}{a^2} - \frac{(x - 6)^2}{b^2} = 1$$ To find a, remember that a is defined as the distance from the center to each vertex. $$\text{Center:} \, (6, -6)$$ $$\text{Vertex:} \, (6, 4)$$ Find the difference in the y-coordinates: $$a = 4 - (-6) = 4 + 6 = 10$$ $$a^2 = (10)^2 = 100$$ $$\frac{(y + 6)^2}{100} - \frac{(x - 6)^2}{b^2} = 1$$ To find b, we use the following relationship. $$b^2 = c^2 - a^2$$ We don't know c yet but it is the distance from the center to each focus. $$\text{Center:} \, (6, -6)$$ $$\text{Focus:} \, (6, -6 + 2\sqrt{34})$$ Find the difference in the y-coordinates: $$c = -6 + 2\sqrt{34} - (-6) = -6 + 2\sqrt{34} + 6 = 2\sqrt{34}$$ $$c^2 = (2\sqrt{34})^2 = (4)(34) = 136$$ $$b^2 = c^2 - a^2$$ $$b^2 = 136 - 100 = 36$$ $$\frac{(y + 6)^2}{100} - \frac{(x - 6)^2}{36} = 1$$

- A hyperbola has two axes of symmetry
- The transverse axis is a line segment that passes through the center and has vertices as its endpoints
- The conjugate axis is a line segment that is perpendicular to the transverse axis and has covertices as its endpoints

- The foci are located on the line that contains the transverse axis
- The transverse and conjugate axes intersect at the center of the hyperbola
- The center of the hyperbola is the midpoint of both the transverse and conjugate axes

- The graph of a hyperbola consists of two disconnected branches that are mirror images of each other
- The points where the branches are closest to the center are the vertices of the hyperbola
- Every hyperbola has two asymptotes that pass through its center
- As we move along each branch away from the center we get closer and closer to the asymptotes
- The fundamental rectangle, centered at the center of the hyperbola, has sides that pass through each vertex and covertex
- The extended diagonals of the fundamental rectangle are the asymptotes of the hyperbola

- The graph of a hyperbola is not the graph of a function
- It does not pass the vertical line test

### Deriving the Standard Form of the Equation of a Hyperbola

We can derive the equation of a hyperbola using its geometric definition. The distance formula for reference: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ The distance from V_{1}(a, 0) to F_{2}(-c, 0): $$d(V_1, F_2) = a - (-c) = a + c$$ The distance from V_{1}(a, 0) to F_{1}(c, 0): $$d(V_1, F_1) = c - a$$ The difference of the distances from the foci to the vertex: $$(a + c) - (c - a) = a + c - c + a = 2a$$ The constant from the definition above is 2a. Instead of using our vertex V_{1}(a, 0), let's apply our definition to any point on the hyperbola P(x, y). $$|d(P, F_2) - d(P, F_1)| = 2a$$ $$\left|\sqrt{(x - (-c))^2 + (y - 0)^2} - \sqrt{(x - c)^2 + (y - 0)^2}\right| = 2a$$ Simplify and apply the definition of absolute value: $$\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm 2a$$ We will be able to achieve the same formula with both +2a and -2a, we will use +2a for the right side. $$\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a$$ Isolate one of the radicals: $$\sqrt{(x + c)^2 + y^2} = 2a + \sqrt{(x - c)^2 + y^2}$$ Square both sides: $$\left(\sqrt{(x + c)^2 + y^2}\right)^2 = \left(2a + \sqrt{(x - c)^2 + y^2}\right)^2$$ $$(x + c)^2 + y^2 = (2a)^2 + (2)(2a)\left(\sqrt{(x - c)^2 + y^2}\right) + (x - c)^2 + y^2$$ Simplify: $$x^2 + 2cx + c^2 + y^2 = 4a^2 + 4a\sqrt{(x - c)^2 + y^2} + x^2 - 2cx + c^2 + y^2$$ $$4cx = 4a^2 + 4a\sqrt{(x - c)^2 + y^2}$$ Divide both sides by 4: $$\require{cancel}\frac{\cancel{4}cx}{\cancel{4}} = \frac{\cancel{4}a^2}{\cancel{4}} + \frac{\cancel{4}a\sqrt{(x - c)^2 + y^2}}{\cancel{4}}$$ $$cx = a^2 + a\sqrt{(x - c)^2 + y^2}$$ Isolate the radical: $$cx - a^2 = a\sqrt{(x - c)^2 + y^2}$$ Square both sides: $$\left(cx - a^2\right)^2 = \left(a\sqrt{(x - c)^2 + y^2}\right)^2$$ Simplify: $$\left(cx - a^2\right)^2 = a^2\left[(x - c)^2 + y^2\right]$$ $$(cx)^2 - 2(cx)(a^2) + (a^2)^2 = a^2(x^2 - 2cx + c^2 + y^2)$$ $$c^2x^2 - 2a^2cx + a^4 = a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2$$ $$c^2x^2 + a^4 = a^2x^2 + a^2c^2 + a^2y^2$$ Rearrange terms: $$c^2x^2 - a^2x^2 - a^2y^2 = a^2c^2 - a^4$$ Factor: $$x^2(c^2 - a^2) - a^2y^2 = a^2(c^2 - a^2)$$ Divide both sides by a^{2}(c^{2}- a^{2}): $$\frac{x^2\cancel{(c^2 - a^2)}}{a^2\cancel{(c^2 - a^2)}} - \frac{\cancel{a^2}y^2}{\cancel{a^2}(c^2 - a^2)} = \frac{1\cancel{a^2(c^2 - a^2)}}{\cancel{a^2(c^2 - a^2)}}$$ $$\frac{x^2}{a^2} - \frac{y^2}{c^2 - a^2} = 1$$ Make a substitution: $$b^2 = c^2 - a^2$$ $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ The above is our equation for a hyperbola with its center at the origin (given as C(0, 0) in the image), with vertices (±a, 0), covertices (0, ±b), and foci (±c, 0). We could use a similar process to derive the standard form of the equation of a hyperbola with its center at the origin and its vertices and foci located on the y-axis.Note: The substitution we made above often causes confusion. Where does b

^{2}= c^{2}- a^{2}come from? After all, in the previous lesson when we derived the equation of an ellipse we saw that we had a point (0, b) that was on the ellipse. We defined b as the distance from the center of the ellipse to each covertex. We were able to use the point (0, b) along with the definition of an ellipse and some basic algebra to prove our relationship in that case. Here, there isn't a point (0, b) on the graph of a hyperbola, and the introduction of b at this point is done purely to make the formula easier to use. We will see later on that b is also defined as the distance from the center to each covertex in a hyperbola. Additionally, it will be used when defining the asymptotes and the fundamental rectangle. Proving the relationship involves creating a right triangle and then using the Pythagorean Theorem. Start with vertices of (±a, 0), and foci of (±c, 0). Form the right triangle with endpoints (0, 0), (a, 0), and (a, b). From the Pythagorean Theorem: $$c^2 = a^2 + b^2$$ Which can be rearranged into: $$b^2 = c^2 - a^2$$ The relationship between a and b can be used to find our asymptotes, which will help us sketch our hyperbola. Here is an article from Math Doctors if you are interested in a deeper discussion of the topic.### Intercepts

Now that we have derived our equation, let's think about the intercepts. $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ To find the x-intercepts, plug in a 0 for y: $$\frac{x^2}{a^2} - \frac{0^2}{b^2} = 1$$ $$\frac{x^2}{a^2} = 1$$ $$x^2 = a^2$$ $$x = \pm a$$ The x-intercepts occur at (±a, 0), which are the same as the vertices in this case.To find the y-intercepts, plug in a 0 for x: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$\frac{0^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$-\frac{y^2}{b^2} = 1$$ $$y^2 = -b^2$$ The above equation does not have any real number solutions, therefore, we will not have any y-intercepts in this case.

### Asymptotes

To find the asymptotes, let's start by solving the equation for y: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$-\frac{y^2}{b^2} = - \frac{x^2}{a^2} + 1$$ $$\frac{y^2}{b^2} = \frac{x^2}{a^2} - 1$$ $$y^2 = \frac{b^2x^2}{a^2} - b^2$$ Before we solve for y, we are going to factor out (b^{2}x^{2})/(a^{2}): $$y^2 = \frac{b^2x^2}{a^2}\left(1 - \frac{a^2}{x^2}\right)$$ Solve for y: $$y = \pm \sqrt{\frac{b^2x^2}{a^2}\left(1 - \frac{a^2}{x^2}\right)}$$ $$y = \pm \sqrt{\frac{b^2x^2}{a^2}} \cdot \sqrt{\left(1 - \frac{a^2}{x^2}\right)}$$ $$y = \pm \sqrt{\left(\frac{bx}{a}\right)^2} \cdot \sqrt{\left(1 - \frac{a^2}{x^2}\right)}$$ $$y = \pm \left|\frac{bx}{a}\right| \cdot \sqrt{\left(1 - \frac{a^2}{x^2}\right)}$$ Since we have the ±, we can drop the absolute value bars. $$y = \pm \frac{b}{a}x \, \sqrt{\left(1 - \frac{a^2}{x^2}\right)}$$ As the |x| approaches positive infinity (|x| → ∞), the fraction a^{2}/x^{2}approaches 0. This means the points on the hyperbola are getting closer and closer to the lines: $$y = \pm \frac{b}{a}x$$ Those lines are known as the asymptotes of the hyperbola and help us sketch the graph. As we move along each branch away from the center the points on the hyperbola get closer and closer to these asymptotes. In other words, these asymptotes are forming a boundary that the hyperbola approaches but never crosses.### Fundamental Rectangle

The fundamental rectangle, which is also known as the central rectangle has its center located at the center of the hyperbola. Its sides pass through each vertex and covertex. We can find the endpoints of the fundamental rectangle as:(a, b), (-a, b), (a, -b), and (-a, -b). The slopes of the diagonals of the fundamental rectangle are ±b/a. We can extend these diagonals to obtain the asymptotes that we found above of y = ±(b/a)x.

### Standard Forms of Equations for Hyperbolas

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$- Center (0, 0)
- Vertices (±a, 0)
- Covertices (0, ±b)
- Foci (±c, 0)
- Asymptotes y = ±(b/a)x
- Endpoints of the fundamental rectangle
- (a, b), (-a, b), (a, -b), (-a, -b)

- Transverse axis is horizontal (lies on the x-axis)
- Conjugate axis is vertical (lies on the y-axis)

- Center (0, 0)
- Vertices (0, ±a)
- Covertices (±b, 0)
- Foci (0, ±c)
- Asymptotes y = ±(a/b)x
- Endpoints of the fundamental rectangle
- (b, a), (-b, a), (b, -a), (-b, -a)

- Transverse axis is vertical (lies on the y-axis)
- Conjugate axis is horizontal (lies on the x-axis)

Example #1: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$\frac{x^2}{9} - \frac{y^2}{25} = 1$$ First, let's match this to our equation from above. We can see the center is at the origin and the transverse axis will lie on the x-axis. $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ $$\frac{x^2}{9} - \frac{y^2}{25} = 1$$ $$a^2 = 9, b^2 = 25$$ $$c^2 = a^2 + b^2$$ $$c^2 = 9 + 25 = 34$$ Note: a, b, and c are positive as defined by our derivation above. Take this into consideration when dealing with square roots in this section. $$a = \sqrt{9} = 3$$ $$\text{Vertices:} \, (3, 0), (-3, 0)$$ $$b = \sqrt{25} = 5$$ $$\text{Covertices:} \, (0, 5), (0, -5)$$ $$c = \sqrt{34}$$ $$\text{Foci:} \, (\sqrt{34}, 0), (-\sqrt{34}, 0)$$ $$\text{Asymptotes:} \, y = \pm \frac{5}{3}x$$ Desmos Link for More Detail

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

$$\frac{y^2}{9} - \frac{x^2}{16} = 1$$

### Translations of Hyperbolas

As we saw with ellipses, the graph of a hyperbola can be centered at (h, k). Horizontal and vertical translations occur when we replace x with (x - h) and y with (y - k) in the standard form of the hyperbola's equation.We can use the following table when considering translations with hyperbolas.

If h and k are positive real numbers (h > 0 and k > 0):

Replacement | Shift |
---|---|

x is replaced with x - h | Shift the graph h units right |

x is replaced with x + h | Shift the graph h units left |

y is replaced with y - k | Shift the graph k units up |

y is replaced with y + k | Shift the graph k units down |

Example #3: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{9} = 1$$ Let's think about matching our form from above. $$1) \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ $$2) \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ Here k = 1, and h = -1, so the center will occur at (-1, 1). In other words, since y has been replaced with (y - 1) and x has been replaced with (x + 1), the center has been shifted 1 unit left and 1 unit up and will now occur at (-1, 1).

The vertices, covertices, and foci are a, b, and c units away from the center, respectively. $$a^2 = 16, a = 4$$ $$b^2 = 9, b = 3$$ $$c^2 = a^2 + b^2 = 16 + 9 = 25, c = 5$$ Since the transverse axis is vertical, to find the vertices, we will start at the center and move up by a units and down by a units. The x-coordinate from the center will not change. $$(-1, 1 + 4) = (-1, 5)$$ $$(-1, 1 - 4) = (-1, -3)$$ $$\text{Vertices:} \, (-1, -3), (-1, 5)$$ Similarly, to find the foci, we will use the same thought process. We will start at the center and move up by c units and down by c units. The x-coordinate from the center will not change. $$(-1, 1 + 5) = (-1, 6)$$ $$(-1, 1 - 5) = (-1, -4)$$ $$\text{Foci:} \, (-1, 6), (-1, -4)$$ To find the covertices we will need to move horizontally since the conjugate axis is horizontal. We will start at the center and move left by b units and right by b units. The y-coordinate from the center will not change. $$(-1 + 3, 1) = (2, 1)$$ $$(-1 - 3, 1) = (-4, 1)$$ $$\text{Covertices:} \, (2, 1), (-4, 1)$$ To find the asymptotes, we will use the point-slope formula. $$y - y_1 = m(x - x_1)$$ Let's go back to our original form for a moment. $$1) \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$ Here, we have asymptotes: y = ±(a/b)x. We know each asymptote passes through the origin, which is the center in this case. $$2) \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ Here, the slope of the asymptotes will remain ±(a/b) and each asymptote will pass through the center, which is (h, k). $$y - y_1 = m(x - x_1)$$ $$y - k = \pm \frac{a}{b}(x - h)$$ $$y = \pm \frac{a}{b}(x - h) + k$$ If we plug in for a, b, h, and k, we get: $$y = \pm \frac{4}{3}(x + 1) + 1$$ If we compare this to: $$y = \pm \frac{4}{3}x$$ We have a shift left by 1 unit coming from (x + 1) and a shift up by 1 unit coming from (+1). This matches the shift from a center of (0, 0) to our current center of (-1, 1).

Asymptote #1 with a slope of 4/3: $$y = \frac{4}{3}(x + 1) + 1$$ $$y = \frac{4}{3}x + \frac{4}{3} + \frac{3}{3}$$ $$y = \frac{4}{3}x + \frac{7}{3}$$ Asymptote #2 with a slope of -4/3: $$y = -\frac{4}{3}(x + 1) + 1$$ $$y = -\frac{4}{3}x - \frac{4}{3} + \frac{3}{3}$$ $$y = -\frac{4}{3}x - \frac{1}{3}$$ The endpoints of the fundamental rectangle can be found as: $$(h + b, k + a) = (-1 + 3, 1 + 4) = (2, 5)$$ $$(h + b, k - a) = (-1 + 3, 1 - 4) = (2, -3)$$ $$(h - b, k + a) = (-1 - 3, 1 + 4) = (-4, 5)$$ $$(h - b, k - a) = (-1 - 3, 1 - 4) = (-4, -3)$$ Desmos Link for More Detail

$$\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{9} = 1$$

### Standard Forms of Equations of Hyperbolas Centered at (h, k)

Using a table to organize the formulas will speed up your work when the center is not at the origin. It is helpful to remember that a is the distance from the center to each vertex, b is the distance from the center to each covertex, and then c is the distance from the center to each focus.Equation | Center | Transverse Axis | Vertices | Covertices | Foci |
---|---|---|---|---|---|

$$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$ | $$(h, k)$$ | Horizontal | $$(h - a, k)$$ $$(h + a, k)$$ | $$(h, k - b)$$ $$(h, k + b)$$ | $$(h - c, k)$$ $$(h + c, k)$$ |

$$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ | $$(h, k)$$ | Vertical | $$(h, k - a)$$ $$(h, k + a)$$ | $$(h - b, k)$$ $$(h + b, k)$$ | $$(h, k - c)$$ $$(h, k + c)$$ |

_{1}, y_{1}) and the slope is ±(b/a) when the transverse axis is horizontal and ±(a/b) when the transverse axis is vertical. We will place this information in the table below along with the endpoints of the fundamental rectangle (F.R.) for quick reference.Equation | Asymptotes | Endpoints F.R. |
---|---|---|

$$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$ | $$y = \pm \frac{b}{a}(x - h) + k$$ | $$(h + a, k + b)$$ $$(h - a, k + b)$$ $$(h + a, k - b)$$ $$(h - a, k - b)$$ |

$$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ | $$y = \pm \frac{a}{b}(x - h) + k$$ | $$(h + b, k + a)$$ $$(h - b, k + a)$$ $$(h + b, k - a)$$ $$(h - b, k - a)$$ |

Horizontal Transverse Axis

Vertical Transverse Axis

Example #4: Find the vertices, covertices, foci, and asymptotes, and then sketch the graph. $$4x^2 - 9y^2 + 8x - 18y - 41 = 0$$ Write in standard form. $$4x^2 + 8x - 9y^2 - 18y - 41 = 0$$ $$4(x^2 + 2x) - 9(y^2 + 2y) = 41$$ $$4(x^2 + 2x + 1 - 1) - 9(y^2 + 2y + 1 - 1) = 41$$ $$4(x^2 + 2x + 1) + (4)(-1) - 9(y^2 + 2y + 1) + (-9)(-1) = 41$$ $$4(x + 1)^2 -4 - 9(y + 1)^2 + 9 = 41$$ $$4(x + 1)^2 - 9(y + 1)^2 + 5 = 41$$ $$4(x + 1)^2 - 9(y + 1)^2 = 36$$ $$\frac{4(x + 1)^2}{36} - \frac{9(y + 1)^2}{36} = \frac{36}{36}$$ $$\frac{(x + 1)^2}{9} - \frac{(y + 1)^2}{4} = 1$$ Match the form and grab the formulas from the table: $$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$ $$h = -1, k = -1$$ $$\text{Center:} \, (-1, -1)$$ $$\text{Vertices:} \, (h - a, k), (h + a, k)$$ $$a^2 = 9, a = 3$$ $$b^2 = 4, b = 2$$ $$h - a = -1 - 3 = -4$$ $$h + a = -1 + 3 = 2$$ $$\text{Vertices:} \, (-4, -1), (2, -1)$$ $$\text{Covertices:} \, (h, k - b), (h, k + b)$$ $$k - b = -1 - 2 = -3$$ $$k + b = -1 + 2 = 1$$ $$\text{Covertices:} \, (-1, -3), (-1, 1)$$ $$\text{Foci:} \, (h - c, k), (h + c, k)$$ $$c^2 = 13, c = \sqrt{13}$$ $$h - c = -1 - \sqrt{13}$$ $$h + c = -1 + \sqrt{13}$$ $$\text{Foci:} \, (-1 - \sqrt{13}, -1), (-1 + \sqrt{13}, -1)$$ $$\text{Asymptotes:} \, y = \pm \frac{b}{a}(x - h) + k$$ $$y = \pm \frac{2}{3}(x + 1) - 1$$ $$y = \frac{2}{3}x + \frac{2}{3} - \frac{3}{3}$$ $$y = \frac{2}{3}x - \frac{1}{3}$$ $$y = -\frac{2}{3}x - \frac{2}{3} - \frac{3}{3}$$ $$y = -\frac{2}{3}x - \frac{5}{3}$$ Endpoints for the Fundamental Rectangle: $$(h + a, k + b) = (-1 + 3, -1 + 2) = (2, 1)$$ $$(h - a, k + b) = (-1 - 3, -1 + 2) = (-4, 1)$$ $$(h + a, k - b) = (-1 + 3, -1 - 2) = (2, -3)$$ $$(h - a, k - b) = (-1 - 3, -1 - 2) = (-4, -3)$$ Desmos Link for More Detail

$$\frac{(x + 1)^2}{9} - \frac{(y + 1)^2}{4} = 1$$

### Finding the Equation of a Hyperbola in Standard Form Given the Vertices and Foci

In some cases, we are given the vertices and foci of the hyperbola and are asked to find the standard form of the equation. We can use the midpoint formula with either the vertices or foci to find the center. Then, we can use the definitions of a and b to find the appropriate denominators for the equation. Let's look at an example.Example #5: Find the standard form of the equation of each hyperbola. $$\text{Vertices:} \, (6, 4), (6, -16)$$ $$\text{Foci:} \, (6, -6 + 2\sqrt{34}), (6, -6 - 2\sqrt{34})$$ Let's start by finding the center. To do this, we will use the midpoint formula with the vertices. $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ $$M = \left(\frac{6 + 6}{2}, \frac{4 + (-16)}{2}\right) = (6, -6)$$ $$\text{Center:} \, (6, -6)$$ Since the y-coordinates of the vertices and foci are different while the x-coordinates remain the same, we know that the transverse axis is vertical. $$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$ Plug in for h and k. Here h = 6 and k = -6. $$\frac{(y + 6)^2}{a^2} - \frac{(x - 6)^2}{b^2} = 1$$ To find a, remember that a is defined as the distance from the center to each vertex. $$\text{Center:} \, (6, -6)$$ $$\text{Vertex:} \, (6, 4)$$ Find the difference in the y-coordinates: $$a = 4 - (-6) = 4 + 6 = 10$$ $$a^2 = (10)^2 = 100$$ $$\frac{(y + 6)^2}{100} - \frac{(x - 6)^2}{b^2} = 1$$ To find b, we use the following relationship. $$b^2 = c^2 - a^2$$ We don't know c yet but it is the distance from the center to each focus. $$\text{Center:} \, (6, -6)$$ $$\text{Focus:} \, (6, -6 + 2\sqrt{34})$$ Find the difference in the y-coordinates: $$c = -6 + 2\sqrt{34} - (-6) = -6 + 2\sqrt{34} + 6 = 2\sqrt{34}$$ $$c^2 = (2\sqrt{34})^2 = (4)(34) = 136$$ $$b^2 = c^2 - a^2$$ $$b^2 = 136 - 100 = 36$$ $$\frac{(y + 6)^2}{100} - \frac{(x - 6)^2}{36} = 1$$

#### Skills Check:

Example #1

Find the standard form of the equation of the hyperbola. $$x^2 - y^2 + 4x - 2y - 78 = 0$$

Please choose the best answer.

A

$$\frac{(y + 1)^2}{81} - \frac{(x + 2)^2}{144} = 1$$

B

$$\frac{(x - 1)^2}{81} - \frac{(y + 2)^2}{121} = 1$$

C

$$\frac{(x + 2)^2}{81} - \frac{(y + 1)^2}{81} = 1$$

D

$$\frac{(y - 2)^2}{36} - \frac{(x - 5)^2}{16} = 1$$

E

$$\frac{(x - 3)^2}{16} - \frac{(y - 1)^2}{64} = 1$$

Example #2

Find the asymptotes. $$-4x^2 + y^2 - 2y - 3 = 0$$

Please choose the best answer.

A

$$y = \pm 3x + \frac{3}{2}$$

B

$$y = \pm \frac{2}{3}x + 1$$

C

$$y = \pm \frac{5}{4}x + 2$$

D

$$y = \pm 2x - 3$$

E

$$y = \pm 2x + 1$$

Example #3

Find the equation of the hyperbola in standard form. $$\text{Vertices:} \, (-10, 6), (-10, -16)$$ $$\text{Foci:} \, (-10, -5 + \sqrt{221}), (-10, -5 - \sqrt{221})$$

Please choose the best answer.

A

$$\frac{(y - 5)^2}{121} - \frac{(x - 10)^2}{100} = 1$$

B

$$\frac{(x - 3)^2}{121} - \frac{(y - 9)^2}{16} = 1$$

C

$$\frac{(y - 12)^2}{20} - \frac{(x - 15)^2}{44} = 1$$

D

$$\frac{(y + 5)^2}{121} - \frac{(x + 10)^2}{100} = 1$$

E

$$\frac{(x + 4)^2}{44} - \frac{(y + 5)^2}{225} = 1$$

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