If We Take Two Identical Cones, One Upright with Its Vertex Pointing Up, Another Upsidedown with Its Vertex Point Down, and Touching the Vertex of the ﬁrst One

Home , Plane curve

If we take two identical cones, one upright with its vertex pointing up, another upsidedown with its vertex point down, and touching the vertex of the first one. Take a plane and cut the cones in different way, and we obtain different goemetric figures. If the plane cuts parallel to the base of one of the cone, the resulting plane curve we get is a circle. If the plane cuts parallel to one of the sides of the cones, the resulting plane curve we get is a parabola. If the plane cuts not parallel to the base of the cone, but intersects only one cone, the resulting plane curve is an ellipse. If the plane cuts through both cones, not parallel to any side or base, the resulting plane curve is a hyperbola.

Parabola Algebraically, A parabola can be defined as the collection of all points (on a plane) that are equidistant from a given point, called the focus and a given line, called the directrix. Note: Remember that by definition, the distance from a point to a line is the perpendicular distance between the point and the line. Example: Let a > 0 be a real number. Find the equation of the parabola whose focus is (0, a) and whose directrix is the line y = a. By definition, we are looking for all the points (x, y) on the plane whose distance (d1) from the point (focus), (0, a), is the same as its distance (d2) from the line (directrix), y = −a. Using the distance formula, we have: p 2 2 d1 = (x − 0) + (y − a) And looking at the picture and knowing that we want the perpendicular distance, we have: d2 = |y − (−a)| = |y + a| (where a is a positive number)

Since we want d1 = d2, we have: p(x − 0)2 + (y − a)2 = |y + a| (x)2 + (y − a)2 = |y + a|2 = (y + a)2 (square both sides) x2 + y2 − 2ay + a2 = y2 + 2ay + a2 (multiply) x2 − 2ay = 2ay (simplify) x2 = 4ay The above example gives us the following: If a > 0, the equation of a parabola with its focus at the point (0, a) and directrix y = −a is given by: 1 4ay = x2 or y = x2 4a Notice this equation contains the origin, (0, 0), and this is the vertex of the parabola. The line that contains the focus and the perpendicular to the directrix, in this case the y−axis, is called the axis of symmetry. The graph of this parabola is a parabola with its vertex at the origin and opens up. Keeping the convention that a > 0, the equation x2 = −4ay gives us a parabola with its directrix being y = a and its focus is the point (0, −a). This is a parabola with vertex at the origin but opens down. The equation y2 = 4ax gives us the equation of the parabola with focus at the point (a, 0) and directrix x = −a. This parabola opens to the right. The equation y2 = −4ax gives us the equation of the parabola with focus at the point (−a, 0) and directrix x = a. This parabola opens to the left. Example: 1 Find the focus, directrix, vertex, and graph the parabola: y = x2 12 Ans: We ﬁrst change it to the form: 4ay = x2, this gives us 12y = x2 ⇒ 4a = 12 ⇒ a = 3. The focus is at the point (0, 3), the directrix is the line y = −3. The vertex of the parabola is at (0, 0). The graph of the parabola opens up.

Example: Find the focus, directrix, vertex, and graph the parabola: −14y = x2. 14 7 Ans: This equation is already in the form we like. We see that a = = , focus 4 2 7 7 is at 0, − , directrix is the line: y = , and vertex is (0, 0). This parabola 2 2 opens down. Example: Find the focus, directrix, vertex, and graph the parabola: 10x = y2 10 5 5 Ans: 4a = 10 ⇒ a = = The focus is on , 0 , directrix is the vertical line 4 2 2 5 x = − , the vertex is still the origin. This parabola opens to the right. 2 Example: Find the focus, directrix, vertex, and graph the parabola: 1 x = − y2 3 3 Ans: Put it into 4ax = y2 form gives −3x = y2. 4a = 3 ⇒ a = . The focus is at 4 3 3 − , 0 , directrix is the vertical line x = . Vertex is the origin. This parabola 4 4 opens to the left. Instead of the origin, if the vertex of the parabola is shifted to another point (h, k), with the directrix still parallel to the x− or y−axis, then the form of the equation that would give us the focus and directrix is: (x − h)2 = 4a(y − k) This is the equation of a parabola whose vertex is at the point (h, k), the directrix is the line y = k − a. Focus is at the point (h, k + a). The axis of symmetry is the line x = h. This parabola opens up. (x − h)2 = −4a(y − k) This is the equation of a parabola whose vertex is at the point (h, k). The directrix is the line y = k + a. Focus is the point (h, k − a). The axis of symmetry is the line x = h. This parabola opens down. 4a(x − h) = (y − k)2 This is the equation of a parabola with vertex at (h, k). The directrix is the line x = h − a. Focus is the point (h + a, k). This parabola opens to the right. −4a(x − h) = (y − k)2 This is the equation of a parabola with vertex at (h, k). The directrix is the line x = h + a. Focus is the point (h − a, k). The parabola opens to the left. When we are given a quadratic expression in standard form, we can turn it into this form by using the method of completing the square. Example: Find the vertex, directrix, focus, and graph the parabola: y = 2x2 − 3x + 1 Ans: We complete the square to turn this into the form we like: 1 3 1 y = x2 − x + 2 2 2 1 1 3 y − = x2 − x 2 2 2 1 1 9 3 9 y − + = x2 − x + 2 2 16 2 16 1 1 32 y + = x − 2 16 4 1 1 32 y + = x − 2 8 4 Comparing to the vertex form 4a(y − k) = (x − h)2, we see that the vertex is 3 1 (h, k) = , − , 4 8 1 1 4a = ⇒ a = 2 8 This is a parabola whose vertex is at 3 1 (h, k) = , − , 4 8 3 the focus is the point (h, k + a) = , 0 . 4 Directrix is the line y = k − a, so directrix is the line: 1 1 1 y = − − = − 8 8 4 3 The axis of symmetry is the line x = 4 Example: Find the vertex, focus, directrix, and graph the parabola x = −4y2 − y + 2 Ans: This time we complete the square on y: x = −4y2 − y + 2 1 1 1 − x = y2 + y − 4 4 2 1 1 1 − x + = y2 + y 4 2 4 1 1 1 1 1 − x + + = y2 + y + 4 2 64 4 64 1 33 12 − x + = y + 4 64 8 1 33 12 − x − = y + 4 16 8 Compare this equation with the vertex form: −4a(x − h) = (y − k)2, we see that the vertex (h, k) is 33 1 (h, k) = , − 16 8 1 1 4a = ⇒ a = 4 16 33 1 1 32 1 1 Focus is the point (h − a, k) = − , − = , − = 2, − 16 16 8 16 8 8 Directrix is the line x = h + a. So directrix is the line: 33 1 34 17 x = + = ⇒ x = 16 16 16 8 1 This is a parabola that opens to the left. Axis of symmetry is the line y = − 8

Example: Find the equation of the parabola with vertex (3, 1) and focus (−2, 1) Ans:

The vertex and the focus lie on the same line of symmetry, y = 1. The distance between the vertex and the focus is a = |3 − (−2)| = 5. Since the focus is on the left hand side of the vertex, this parabola opens to the left. Its equation is: −4(a)(x − h) = (y − k)2. Since a = 5 and (h, k) = (3, 1), the equation of the parabola is: −4(5)(x − 3) = (y − 1)2 −20(x − 3) = (y − 1)2 The directrix is on the other side of the vertex opposite the focus. The equation of the directrix is x = 8.

Example: Find the equation of the parabola with focus (−2, −1) and directrix y = 3. Ans:

The axis of symmetry is perpendicular to the directrix. This means the axis of symmetry is parallel to the x−axis, meaning the vertex with have the same x−coordinate as the focus. The vertex of the parabola is halfway between the focus and the directrix, so the vertex is (−2, 1) The distance between the focus and the vertex is a = |1 − (−1)| = 2. This is a parabola that opens up. The general equation is: −4a(y − k) = (x − h)2. Since a = 2 and (h, k) = (−2, 1), the equation for this parabola is: −4(2)(y − 1) = (x − (−2))2 −8(y − 1) = (x + 2)2

Example: Find the equation of the parabola given by the graph. Ans: We see from the graph that the parabola has vertex (1, −3). This is a parabola that opens to the right, so its equation will be of the form: 4a(x − h) = (y − k)2 Knowing the vertex is (1, −3) allows us to write the equation as: 4a(x − 1) = (y − (−3))2 4a(x − 1) = (y + 3)2 5 To ﬁnd the value of a, we use the additional point, , 0 from the graph. When 2 5 x = , y = 0, we have: 2 5 4a − 1 = (0 + 3)2 2 3 4a = (3)2 2 6a = 9 9 3 a = = 6 2 Finally, the equation of the parabola is: 3 4 (x − 1) = (y + 3)2 2 6 (x − 1) = (y + 3)2 Circles Deﬁnition: A circle is the collection of all points (on a plane) equidistant from a given point called the center of the circle. The distance between the center and any point on the circle is the radius of the circle. A line that contains the center and connects two points on (opposite ends of) the circle is a diameter of the circle.

Example: Find the equation of the circle centered at the origin and has radius r

Answer: Let (x, y) be any point on the circle. Using Pythegorean Theorems, we see that x2 + y2 = r2. This is the standard equation of the circle centered at the origin with radius r.

Example: Find the equation of the circle centered at the origin with radius 3. Ans: Using the standard form, we have: x2 + y2 = 9. Example: Find the center and radius of the circle: x2 + y2 = 15 √ Ans: The center is at the origin. Radius is r = 15

If the center of the circle is (h, k), then the equation of the circle with radius r and center at (h, k) is given by: (x − h)2 + (y − k)2 = r2 Example: Find the equation of the circle with center (2, −3) and radius r = 2. Ans: Putting the numbers into the standard equation, we have: (x − 2)2 + (y − (−3))2 = 22 (x − 2)2 + (y + 3)2 = 4

Example: Find the equation of the circle with center at (−1, 1) and contains the point (1, −2). Ans: We first need to find the radius of the circle. The radius of the circle is the distance between the center of the circle to any point on the circle. Using the distance formula, we have: √ √ r = p(1 − (−1))2 + (−2 − 1)2 = 4 + 9 = 13 The equation of the circle is: (x + 1)2 + (y − 1)2 = 13 Example: Find the center and radius of the circle, and graph the equation: x2 − 5x + y2 + 6y = 0 Ans: To put this in standard form, we need to first complete the square: 25 25 x2 − 5x + + y2 + 6y + 9 = 0 + + 9 4 4 52 61 x − + (y + 3)2 = 2 4 √ 5 61 Center of circle is , −3 , r = 2 2

Ellipse An ellipse is the collection of points on a plane the sume of whose distance from two ﬁxed points, called the foci, is a constant. To draw an ellipse, imagine you took a string and pined the two end points of the string at a distance less than the length of the string, then take a pencil and trace along the string, you will get the graph of an ellipse. The location of the two end points where the string is ﬁxed is the foci, and the length of the string is the constant length.

In an ellipse, the line that contains the two foci is the major axis. The midpoint between the two foci is the center of the ellipse, and the line that goes through the center and is perpendicular to the major axis is the minor axis. The two points on the ellipse where the major axis and the ellipse intersect are the vertices of the ellipse. Example: Find the equation of the ellipse with foci at (−c, 0) and (c, 0), and the sum of the distance between any point of the ellipse and the two foci is given by 2a.

Ans: This is an ellipse whose major axis is the x−axis and its center at the origin. Let (x, y) be any point on the ellipse, let 2a be the sum of the distance from (x, y) to the two points (using 2a instead of a makes the formula easier to derive and represent). With the distance formula we have: p(x − (−c))2 + (y − 0)2 + p(x − c)2 + (y − 0)2 = 2a Moving the second radical term to the right gives: p(x + c)2 + (y)2 = 2a − p(x − c)2 + (y)2 Square both sides gives: (x + c)2 + y2 = 4a2 − 4ap(x − c)2 + y2 + (x − c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 − 4ap(x − c)2 + y2 + x2 − 2cx + c2 + y2 Cancel common terms on both sides: 2cx = 4a2 − 4ap(x − c)2 + y2 + −2cx Isolating the remaining radical: 4cx − 4a2 = −4ap(x − c)2 + y2 Divide both sides by 4: cx − a2 = −ap(x − c)2 + y2 Square both sides again: (cx − a2)2 = a2((x − c)2 + y2) Multiply then simplify: c2x2 − 2a2cx + a4 = a2(x2 − 2cx + c2 + y2) c2x2 − 2a2cx + a4 = a2x2 − 2a2cx + a2c2 + a2y2 c2x2 + a4 = a2x2 + a2c2 + a2y2 Rearrange the terms: a4 − a2c2 = a2x2 − c2x2 + a2y2 Factor: a2(a2 − c2) = (a2 − c2)x2 + a2y2 Notice from the picture of the ellipse that a > c because of the triangle inequality, so a2 − c2 > 0. Let b be the positive number such that b2 = a2 − c2, then the above equation can be simpliﬁed as: a2b2 = b2x2 + a2y2 Divide both sides of the equation by a2b2 gives us: a2b2 b2x2 + a2y2 b2x2 a2y2 = = + a2b2 a2b2 a2b2 a2b2 x2 y2 1 = + a2 b2 The above example gives us the Standard Equation of an Ellipse with center at the origin: x2 y2 + = 1 a2 b2 In the above equation, a, b are both positive numbers. If a > b, then the x−axis is the major axis and y − axis is the minor axis. The foci will be on the x−axis and have the coordinates (−c, 0) and (c, 0), where c2 = a2 − b2. The length of the major axis is equal to 2a, and the length of the minor axis is 2b. The vertices are (−a, 0) and (a, 0) If b > a, then the major axis is the y−axis and the minor axis is the x−axis. The foci will be on the y−axis and will have coordinates (0, −c) and (0, c), where c2 = b2 − a2. The length of the major axis is 2b, and length of minor axis is 2a. The vertices ar (0, −b) and (0, b) If a = b, we have a circle. Example: Find the foci, major and minor axis and graph the ellipse: x2 y2 + = 1 49 25 Ans: In the above equation, a = 7, b = 5, since a > b, the major axis is on the x−axis. The length of the major axis is 2(7) = 14. The length of the minor axis is 2(5) = 10. The foci are on the major axis. Letting c2 = a2 − b2 gives us: √ c2 = 49 − 25 = 24 ⇒ c = 24 The coordinates of the two foci are: √ √ (− 24, 0), ( 24, 0) The vertices are (7, 0) and (−7, 0) Example: Find the foci, major and minor axis, and graph the ellipse: x2 y2 + = 1 3 5 √ √ Ans: In this example, a = 3, b =√ 5. Since b > a, the major axis is on the y−axis. Length√ of major axis is 2 5. Minor axis is on the x−axis, length of minor axis is 2 3. The foci are on the major (y−)axis. Letting c2 = b2 − a2 gives: √ c2 = 5 − 3 = 2 ⇒ c = 2 The coordinates of the two foci are: √ √ (0, − 2), (0, 2) √ √ Vertices are (0, − 5), (0, 5) Example: Find the foci, major and minor axis, and graph the ellipse: 4x2 + 10y2 = 1 Ans: This equation is the equation of an ellipse. We want to write this in the x2 y2 form of + = 1 a2 b2 1 We use the fact that for any number n, n = 1/n 1 1 So 4 = and 10 = , we have: 1/4 1/10 4x2 + 10y2 = 1 x2 y2 + = 1 1/4 1/10 1 1 1 1 Therefore, a2 = ⇒ a = , b2 = ⇒ b = √ 4 2 10 10 1 Since a > b, the major axis is on the x−axis. Length of major axis is 2 = 1, 2 1 2 length of minor axis = 2 √ = √ 10 10 √ 1 1 3 3 Let c2 = a2 − b2 = − = ⇒ c = √ 4 10 20 20 √ ! √ ! 3 3 Foci are at −√ , 0 , √ , 0 20 20 1 1 Vertices at − , 0 , , 0 2 2

Example: Find the equation of the ellipse with center at the origin, one of the foci at (3, 0) and a vertex at (4, 0). Ans:

Since the focus is on the x−axis, the major axis is on the x−axis. c = 3. a = 4. √ b2 = a2 − c2 = 42 − 32 = 16 − 9 = 7 ⇒ b = 7 The equation of the ellipse is: x2 y2 + = 1 16 7 The standard equation of an ellipse with its center at (h, k), major and minor axis still parallel to x− and y−axis, is given by: (x − h)2 (y − k)2 + = 1 a2 b2 If a > b, the major axis will be parallel to the x−axis, and on the line y = k. The foci will have coordinates (h − c, k), (h + c, k), where c is given by c2 = a2 − b2. The vertices are (h − a, k), (h + a, k). If b > a, the major axis will be parallel to the y−axis, and on the line x = h. The foci will have coordinates (h, k−c), (h, k+c), where c is given by c2 = b2 −a2 = c2 The vertices are (h, k − a), (h, k + a). In the standard equation of an ellipse, suppose a > b (so length of major axis =2a), we deﬁne the eccentricity, e, of an ellipse by c e = a The number e is always between 0 and 1 for an ellipse. e measures how close to a circle or a straight line an ellipse is. The closer e is to 0, the more rounded an ellipse is. The closeer e is to 1, the more ﬂat or close to a straight line an ellipse is. Example: Find the center, foci, major and minor axis, and graph the ellipse: (x − 2)2 (y + 1)2 + = 1 6 16 √ Ans: The center of this ellipse is at the point (2, −1). a = 6, b = 4. Since b > a, the major axis is parallel to the y−axis. The equation of the major axis is x = 2. Equation of the minor axis is y = −1. √ Let c2 = b2 − a2 = 16 − 6 = 10 ⇒ c = 10 √ √ The foci are the two points (2, −1 − 10), (2, −1 + 10) Vertices are (2, −5), (2, 3) Example: Find the center, foci, major and minor axis, and graph the ellipse: 6x2 + 2y2 − 12x + 4y + 1 = 0 Ans: This equation is not yet in standard form. We need to complete the square on both x and y to turn this equation to standard form of an ellipse: 6x2 − 12x + 2y2 + 4y + 1 = 0 (6x2 − 12x) + (2y2 + 4y) = −1 6(x2 − 2x) + 2(y2 + 2y) = −1 6(x2 − 2x + 1) + 2(y2 + 2y + 1) = −1 + 6(1) + 2(1) 6(x − 1)2 + 2(y + 1)2 = 7 6(x − 1)2 + 2(y + 1)2 = 1 7 6(x − 1)2 2(y + 1)2 + = 1 7 7 (x − 1)2 (y + 1)2 + = 1 7/6 7/2 √ √ 7 7 So the center is at (1, −1). a = √ , b = √ . Since b > a, the major axis is 6 2 parallel to the y-axis. The major axis has the equation x = 1. The minor axis is parallel to the x-axis and has equation y = −1. √ 7 7 14 7 7 Let c2 = b2 − a2 = − = = ⇒ c = √ 2 6 6 3 3 √ ! √ ! 7 7 The foci are at 1, −1 − √ , 1, −1 + √ . 3 3 √ ! √ ! 7 7 Vertices are 1, −1 − √ , 1, −1 + √ 2 2

Example: Find the equation of the ellipse with foci at (−1, 2) and (3, 2), and the length of major axis is 7 Ans: The center is halfway between the two foci, so center is the point (1, 2). The foci have the same y coordinate, so the major axis is parallel to the x axis. The major axis is the line y = 2, minor axis is the line x = 1. 7 The length of the major axis is equal to 2a, so a = . 2 7 7 Vertices at 1 − , 2 and 1 + , 2 ⇒ 2 2 5 9 Vertices at − , 2 and , 2 2 2 The distance from the center to one of the foci is c, so c = |3 − 1| = 2 √ 49 49 16 33 33 b2 = a2 − c2 = − 4 = − = ⇒ b = . 4 4 4 4 2 Putting all these togehter gives: (x − 1)2 (y − 2)2 2 + √ 2 = 1 (7/2) 33/2 (x − 1)2 (y − 2)2 + = 1 49/4 33/4 Hyperbola Definition: A hyperbola is the collection of all points in a plane the difference of the distances from two fixed points, called the foci, is a constant. Graphically, a hyperbola is consisted of two pieces of symetric curves. The line that contains the foci is called the transverse axis. The midpoint between the foci is the center. The intersections between the transverse axis and the hyperbola are the vertices of the hyperbola. The line through the center and perpendicular to the transverse axis is the conjugate axis.

Example: Find the equation of the hyperbola with center at the origin and foci at the points (−c, 0) and (c, 0), and the constant diﬀerence of the distance is equal to 2a. Let (x, y) be any point on the hyperbola. We want the (absolute) diﬀerence of distance between this point and the two foci to be 2a. Using the distance formula, we have the equation:

|d1 − d2| = 2a d1 − d2 = ±2a p(x − (−c))2 + (y − 0)2 − p(x − c)2 + (y − 0)2 = ±2a p(x + c)2 + y2 − p(x − c)2 + y2 = ±2a p(x + c)2 + y2 = ±2a + p(x − c)2 + y2 Square both sides: (x + c)2 + y2 = 4a2 ± 4ap(x − c)2 + y2 + (x − c)2 + y2 Multiply then simplify: x2 + 2cx + c2 + y2 = 4a2 ± 4ap(x − c)2 + y2 + x2 − 2cx + c2 + y2 2cx = 4a2 ± 4ap(x − c)2 + y2 + −2cx Isolating the radical expression and simplify: 4cx − 4a2 = ±4ap(x − c)2 + y2 cx − a2 = ±ap(x − c)2 + y2 Square both sides again: (cx − a2)2 = a2 (x − c)2 + y2 Multiply and simplify: c2x2 − 2cxa2 + a4 = a2 x2 − 2cx + c2 + y2 c2x2 − 2cxa2 + a4 = a2x2 − 2cxa2 + a2c2 + a2y2 c2x2 + a4 = a2x2 + a2c2 + a2y2 Rearrange the terms and factor: c2x2 − a2x2 − a2y2 = a2c2 − a4 (c2 − a2)x2 − a2y2 = a2(c2 − a2) Using the triangle inequality, one can see that c > a, so c2 − a2 > 0. Let b > 0 such that b2 = c2 − a2, then: b2x2 − a2y2 = a2b2 Multiply both sides of the equation by a2b2 gives: b2x2 a2y2 a2b2 − = a2b2 a2b2 a2b2 x2 y2 − = 1 a2 b2 We have discovered the: Standard Equation of a Hyperbola with Center at the Origin, foci on the x-axis: x2 y2 − = 1 a2 b2 For this hyperbola, the vertices are at (−a, 0) and (a, 0) (on the x−axis); the foci are the points (−c, 0) and (c, 0) (on the x−axis), where c2 = a2 +b2. Graphically, this parabola opens left/right. Standard Equation of a Hyperbola with Center at the Origin, foci on the y-axis: x2 y2 − + = 1 a2 b2 For this hyperbola, the vertices are at (0, −b) and (0, b) (on the y−axis); the foci are the points (0, −c) and (0, c) (on the y−axis), where c2 = a2 + b2. Graphically, this parabola opens up/down. If we solve for y in the above equation, we get: y2 x2 = + 1 b2 a2 x2 y2 = b2 + 1 a2 b2 y2 = x2 + b2 a2 Completely factor out the ﬁrst term on the right gives: b2 a2 y2 = x2 1 + a2 x2 s b a2 y = ± x 1 + a x2 Notice that as the value of x becomes extremely large (x approaches inﬁnity, s a2 a2 x → ∞), the fraction will appraoch 0. So the expression 1 + will x2 x2 √ √ b approach 1 + 0 = 1 = 1. This tells us that, as x → ∞, y ≈ ± x. a In other words, in the equation above, as x becomes extremely large, the value b of y will be closely approximated by the two lines y = ± x. a We call the two lines asymptotes of the parabola. An asymptote to a graph is a straight line that the graph will get arbitrarily close to.

Theorem: x2 y2 x2 y2 The hyperbolas − = 1 and the hyperbolas − + = 1 have two (slant) a2 b2 a2 b2 asymptotes given by the equations: b b y = x and y = − x a a

Example: Find the foci, vertices, asymptotes, and graph the hyperbola: x2 y2 − = 1 9 16 Ans: This is a hyperbola with center at the origin. The positive sign is on x2. This tells us that the foci are on the x−axis, and the parabola opens to the side. a = 3, b = 4, solving for c give: c2 = a2 + b2 = 9 + 16 = 25 ⇒ c = 5. The foci are at the points (−5, 0), (5, 0). The vertices are at (a, 0) = (−3, 0), (−a, 0) = (3, 0) The asymptotes of the hyperbola are the two lines: 4 4 y = and y = − 3 3

Example: Find the foci, vertices, asymptotes, and graph the hyperbola: −x2 + 6y2 = 1 Ans: Rewrite the equation as: y2 −x2 + = 1 1/6 1 So a = 1, b = √ . 6 Since the positive sign is on y, this hyperbola opens up/down. The foci are on the y−axis. √ 1 7 7 c2 = a2 + b2 = 1 + = ⇒ c = √ 6 6 6 √ ! √ ! 7 7 The foci are at the points 0, −√ , 0, √ . 6 6 1 The vertices are on the y−axis. They have coordinates (0, −b) = 0, −√ , 6 1 (0, b) = 0, √ . 6 b The asymptotes are y = ± x. In this example, they are: a √ √ 1/ 6 1 1/ 6 1 y = − x = −√ x and y = x = √ x 1 6 1 6 If the center of the hyperbola is moved to a new point (h, k), but the trasverse is still parallel to the x−axis or y−axis, then Standard Equation of Hyperbola with Center at (h, k), transverse parallel to x-axis: (x − h)2 (y − k)2 − = 1 a2 b2 For this hyperbola, the center is at (h, k). Since positive sign is on x, the foci are on a transverse parallel to the x−axis. Coordinate of foci are (h−c, k), (h+c, k), where c2 = a2 + b2. Vertices have coordinate (h − a, k), (h + a, k). b Asymptotes have equation: y − k = ± (x − h) a Standard Equation of Hyperbola with Center at (h, k), transverse parallel to y-axis: (x − h)2 (y − k)2 − + = 1 a2 b2 For this hyperbola, the center is at (h, k). Since positive sign is on y, the foci are on a transverse parallel to the y−axis. Coordinate of foci are (h, k −c), (h, k +c), where c2 = a2 + b2. Vertices have coordinate (h, k − a), (h, k + a). b Asymptotes have equation: y − k = ± (x − h) a

Example: Find the vertices, foci, asymptotes, and graph the parabola: (x − 2)2 (y + 4)2 − = 1 10 15 Ans: Center√ is at (2√, −4). Since positive sign on x, transverse is parallel to x axis. a = 10, b = 15. c2 = a2 + b2 = 10 + 15 = 25 ⇒ c = 5 Foci at (2 − 5, −4) = (−3, −4), (2 + 5, −4) = (7, −4) √ √ Vertices at (2 − 10, −4), (2 + 10, −4) Asymptotes are 15 3 y − (−4) = − (x − 2) ⇒ y + 4 = − (x − 2) and 10 2 15 3 y − (−4) = (x − 2) ⇒ y + 4 = (x − 2) 10 2 Example: Find the vertices, foci, asymptotes, and graph the hyperbola: −8x2 − 16x + 3y2 + 18y − 2 = 0 Ans: We need to complete the square to turn the equation into standard form: −8(x2 + 2x) + 3(y2 + 6y) = 2 −8(x2 + 2x + 1) + 3(y2 + 6y + 9) = 2 + (−8)(1) + (3)(9) −8(x + 1)2 + 3(y + 3)2 = 21 −8(x + 1)2 3(y + 3)2 + = 1 21 21 −8 (y + 3)2 (x + 1)2 + = 1 21 7 (x + 1)2 (y + 3)2 − + = 1 21/8 7 r21 √ Center = (−1, −3). a = , b = 7 8 Since positive sign on y, transverse is parallel to y−axis. Foci on y−axis. 21 77 r77 c2 = a2 + b2 = + 7 = ⇒ c = 8 8 8 ! ! r77 r77 Foci at −1, −3 − , −1, −3 + 8 8 √ √ Vertices at −1, −3 − 7 , −1, −3 + 7 √ √ 7 8 Asymptotes are: y − (−3) = −p (x − (−1)) ⇒ y + 3 = −√ (x + 1) and √ 21/8 √ 3 7 8 y − (−3) = (x − (−1)) ⇒ y + 3 = √ (x + 1) p21/8 3

Example: Find the equation of the hyperbola with its center at (1, 4), a focu at (−2, 4), and a vertex at (0, 4). Ans: The center and focu have the same y−coordinate, so the transverse is parallel to the x−axis. In other words, the positive sign will be with x in the equation. The distance between the foci and the center is c = |1 − (−2)| = 3. Since the center is halfway between the two foci, the other focus is at (4, 4) The distance between the center and the vertex is |1 − 0| = 1 = a. c2 = a2 + b2 ⇒ 32 = 12 + b2 ⇒ b2 = 8 The equation of the hyperbola is: (x − 1)2 (y − 4)2 − = 1 1 8 (y − 4)2 (x − 1)2 − = 1 8 √ The equations of the asymptotes are y − 4 = ± 8(x − 1)

Example: Find the equation of the hyperbola with vertices at (−1, 5) and (−1, −1), and one of its asymptotes has equation y − 2 = 2(x + 1) Ans: Since the two vertices have the same x−coordinate, the transverse is parallel to the y−axis. This means the positive sign will be with y in the equation. The center is halfway between the two vertices, so center = (−1, 2) The distance between the center and one of the vertex is b = |2 − 5| = 3 b The equation of an asymptote of the hyperbola has the form: y − k = (x − h). a Comparing the general equation with y − 2 = 2(x + 1) we see that b b 3 = 2 ⇒ a = = a 2 2 √ 32 9 45 45 c2 = a2 + b2 = + (3)2 = + 9 = ⇒ c = 2 4 4 2 √ ! √ ! 45 45 Foci at −1, 2 − , −1, 2 + 2 2 Equation of the other asymptote is y − 2 = −2(x + 1) The equation of the hyperbola is: (x − (−1))2 (y − 2)2 − + = 1 (3/2)2 32 (x + 1)2 (y − 2)2 − + = 1 9/4 9 4(x + 1)2 (y − 2)2 − + = 1 9 9