The Hyperbola Week 18

Definition of a hyperbola The locus of a point which moves in such a way that the ratio of its distance from a fixed point (focus) and from a fixed straight line (directrix) is a constant greater than 1 is a hyperbola.

The Equation of a Hyperbola S is the focus ZL is the directrix A is a point on the hyperbola O is the mid-point of AA’ SA = e.AZ (e>1) P is another point on the hyperbola L N P(x, y) A’ O Z A S

The Equation of a Hyperbola SA= e.AZ (e>1) SA’=e.A’Z SA’-SA=e(A’Z-AZ) (OS+a)-(OS-a)=e(A’Z-AZ) 2a=e[(a+OZ)-(a-OZ)] 2a=2e.OZ OZ=a/e (Important) L N P(x, y) A’ O Z A S

The Equation of a Hyperbola (OS-a)=SA (OS-a)=eAZ OS-a=e(a-OZ) OS=a+e(a-a/e) OS=a+ae-a OS=ae (Important) L N P(x, y) A’ O Z A S

The Equation of a Hyperbola P(x, y) A’ O Z A S PS = e.PN (e>1) PS2=e2PN

Alternative Definition The set of all points in a plane such that the difference of the distances to 2 fixed points (foci) is constant. Standard form of a hyperbola centered at the origin: Opens left & right Opens up & down

What are Hyperbolas? Hyperbolas can be thought of as two parabolas going in opposite directions

Hyperbola F1 F2 d1 d2 P For any point P that is on the hyperbola, d2 – d1 is always the same (is constant). In this example, the origin is the center of the hyperbola. It is midway between the foci.

Hyperbolas F V C 1. A line through the foci intersects the hyperbola at two points, called the vertices (V) 2. The segment connecting the vertices is called the transverse axis of the hyperbola. 3. The center of the hyperbola is located at the midpoint of the transverse axis.

Hyperbolas F V C As x and y get larger the branches of the hyperbola approach a pair of intersecting lines called the asymptotes of the hyperbola. These asymptotes pass through the center of the hyperbola.

The Standard Equation of a Hyperbola With Center (0, 0) and Foci on the x-axis The equation of a hyperbola with the centre (0, 0) and foci on the x-axis is: B (0, b) The length of the transverse axis is 2a. The length of the conjugate axis is 2b. The vertices are (a, 0) and (-a, 0). The foci are (c, 0) and (-c, 0). The slopes of the asymptotes are (-c, 0) A1 A2 (c, 0) F1 (-a, 0) (a, 0) F2 B (0, -b) The equations of the asymptotes

The Standard Equation of a Hyperbola with Center (0, 0) and Foci on the y-axis The equation of a hyperbola with the centre (0, 0) and foci on the y-axis is: F1(0, c) The length of the transverse axis is 2a. The length of the conjugate axis is 2b. The vertices are (0, a) and ( 0, -a). The foci are (0, c) and (0, -c). The slopes of the asymptotes are A1(0, a) B1(-b, 0) B2(b, 0) A2(0, -a) F2(0, -c) The equations of the asymptotes

State the coordinates of the vertices, the coordinates of the foci, Hyperbola Properties State the coordinates of the vertices, the coordinates of the foci, the lengths of the transverse and conjugate axes, and the equations of the asymptotes of the hyperbola defined by each equation. The equations of the asymptotes are For this equation, a = 2 and b = 4. The length of the transverse axis is 2a = 4. The length of the conjugate axis is 2b = 8. The vertices are (2, 0) and (-2, 0): c2 = a2 + b2 = 4 + 16 = 20 The coordinates of the foci are

Hyperbola Properties For this equation, a = 5 and b = 3. The length of the transverse axis is 2a = 10. The length of the conjugate axis is 2b = 6. The vertices are (0, 5) and (0, -5): c2 = a2 + b2 = 25 + 9 = 34 The coordinates of the foci are The equations of the asymptotes are

The Standard Form of the Hyperbola with Centre (h, k) When the transverse axis is vertical, the equation in standard form is: The centre is (h, k). The transverse axis is parallel to the y-axis and has a length of 2a units. The conjugate axis is parallel to the x-axis and has a length of 2b units. The slopes of the asymptotes are (h, k) The general form of the equation is Ax2 + Cy2 + Dx + Ey + F = 0.

The Standard Form of the Hyperbola with Centre (h, k) [cont’d] When the transverse axis is horizontal, the equation in standard form is: The transverse axis is parallel to the x-axis and has a length of 2a units. The conjugate axis is parallel to the y-axis and has a length of 2b units. The slopes of the asymptotes are

Describe the hyperbola Transverse axis is vertical Centered at (-1,3) Distance to vertices from center = 2 units (up & down) (-1,5) & (-1,1) Asymptotes pass through the (-1,3) with slopes = 2/3, -2/3 Foci units up & down from the center ,

Finding the Equation of a Hyperbola The centre is (2, 3), so h = 3 and k = 2. The transverse axis is parallel to the y-axis and has a length of 10 units, so a = 5. The conjugate axis is parallel to the x-axis and has a length of 6 units, so b = 3. The vertices are (-2, 8) and (-2, -2). The slope of one asymptote is , so a = 5 and b = 3: c2 = a2 + b2 = 25 + 9 = 34 The coordinates of the foci are Standard form

Writing the Equation in its General Form 9(y - 3)2 - 25(x +2)2 = 225 9(y2 - 6y + 9) - 25(x2 + 4x + 4) = 225 9y2 - 54y + 81 - 25x2 - 100x - 100 = 225 -25x2 + 9y2 - 100x - 54y + 81 - 100 = 225 -25x2 + 9y2 - 100x - 54y - 244 = 0 The general form of the equation

Writing the Equation of a Hyperbola Write the equation of the hyperbola with centre at (2, -3), one vertex at (6, -3), and the coordinates of one focus at (-3, -3). The centre is (2, -3), so h = 2, k = -3. The distance from the centre to the vertex is 4 units, so a = 4. The distance from the centre to the foci is 5 units, so c = 5. Use the Pythagorean property to find b: b2 = c2 - a2 = 25 - 16 = 9 b = ± 3 9(x - 2)2 - 16(y + 3)2 = 1 9(x2 - 4x + 4) - 16(y2 + 6y + 9) = 144 9x2 - 36x + 36 - 16y2 - 96y - 144 = 144 9x2 - 16y2 - 36x - 96y + 36 - 144 = 144 9x2 - 16y2 - 36x - 96y - 216 = 0 General form Standard form

State the coordinates of the vertices, the coordinates of Sample Problem State the coordinates of the vertices, the coordinates of the foci, the lengths of the transverse and conjugate axes and the equations of the asymptotes of the hyperbola defined by 4x2 - 9y2 + 32x + 18y + 91 = 0. 4x2 - 9y2 + 32x + 18y + 91 = 0 (4x2 + 32x ) + (- 9y2 + 18y) + 91 = 0 4(x2 + 8x + ____) - 9(y2 - 2y + _____) = -91 + _____ + _____ 16 1 64 -9 4(x + 4)2 - 9(y - 1)2 = -36 3.5.14

General Information about Hyperbolas Centered at Origin x and y are both squared Equation always equals(=) 1 Equation is always minus(-) a2 is always the first denominator c2 = a2 + b2 c is the distance from the center to each foci on the major axis a is the distance from the center to each vertex on the major axis

More General Information about Hyperbolas Centered at the Origin b is the distance from the center to each midpoint of the rectangle used to draw the asymptotes. This distance runs perpendicular to the distance (a). Major axis has a length of 2a Eccentricity(e): e = c/a (The closer e gets to 1, the closer it is to being circular If x2 is first then the hyperbola is horizontal If y2 is first then the hyperbola is vertical.

What changes if the center is NOT at the origin? Standard Form of Equation Equations of Asymptotes

Essential Properties of Hyperbolas Equation of Hyperbola Center Foci Points Equation of Asymptote Vertex Transverse Axis (x – h)2 _ (y – k)2 a2 b2 (h, k) (h – c, k) and (h + c, k) y – k = +/- (b/a) (x – h (h +a, k) and (h – a, k) Horizontal (y – k)2 _ (x – h)2 a2 b2 (h, k – c) and (h, k + c) (c = a2 + b2 ) y – k = +/- (a/b) (x – h) (h, k + a) and (h, k – a) Vertical = 1 = 1 You must be looking at a hyperbola when the x2 and y2 terms are subtracted (x2 – y2) or (y2 – x2)

Practice Example 1 Write the standard form of the equation of the hyperbola 144y2 – 25x2 – 576y – 150x = 3249. Then find the coordinates of the center, the vertices, the foci, and the equation of the asymptotes. Graph the hyperbola and the asymptotes.

Practice Example 1 Write the standard form of the equation of the hyperbola 144y2 – 25x2 – 576y – 150x = 3249. Then find the coordinates of the center, the vertices, the foci, and the equation of the asymptotes. Graph the hyperbola and the asymptotes. 144(y2 – 4y + o) – 25(x2 + 6x + o) = 3249 + 144(o) + 25(o) 144(y2 – 4y + 4) – 25(x2 + 6x + 9) = 3249 + 144(4) + 25(9) 144(y – 2)2 – 25(x + 3)2 = 3600 (y-2)2 _ (x + 3)2 25 144 = 1 Center: (-3, 2) a = 5 so the vertices are (-3, 7) and (-3, -3) a2 + b2 = c2 25 + 144 = c2 c = 13 The foci are (-3, 15) and (-3, -11).

Practice Example 1 (cont.) Asymptotes have the formula y = +/- a/b x and we have center (-3, 2) and slopes +/- 5/12. y – 2 = 5/12 (x + 3) y – 2 = -5/12 (x + 3) y – 2 = (5/12) x + 15/12 y – 2 = (-5/12) x + -15/12 y = (5/12) x + 13/4 y = (-5/12) x + 3/4

Practice Problem 2 Find the coordinates of the vertices and the foci. Give the asymptote slopes for each hyperbola. Then draw the graph. x2 _ y2 9 49 2) 25x2 – 4y2 = 100 = 1 Vertices: (-2, 0) (2, 0) Foci: ( 29, 0) (- 29, 0) Slope = +/- 5/2 Vertices: (-3, 0) (3, 0) Foci: (- 58, 0) ( 58, 0) Slope = +/- 7/3 2) 1)

Further Problems Identify 9x2 + 16y2 – 54x + 64y + 1 = 0 as one of the four conic sections. Then graph the conic section. 9x2 + 16y2 – 54x + 64y = -1 9 (x2 – 6x + o) + 16(y2 + 4y + o) = -1 + 9(o) + 16(o) 9 (x2 – 6x + 9) + 16(y2 + 4y + 4) = -1 + 9(9) + 16(4) 9(x – 3)2 + 16(y + 2)2 = 144 (x – 3)2 (y + 2)2 9 This conic section is an ellipse. + = 1

Further Problems Identify 9x2 + 16y2 – 54x + 64y + 1 = 0 as one of the four conic sections. Then graph the conic section.

1) (x + 3)2 + (y)2 = 16 circle 2) (x – 3)2 _ (y + 1)2 hyperbola Further Problems Write the equation in standard form and decide if the conic section is a parabola, a circle, an ellipse, or a hyperbola. Then graph the equation. x2 + y2 + 6x = 7 5x2 – 6y2 – 30x – 12y + 9 = 0 6 5 1) (x + 3)2 + (y)2 = 16 circle 2) (x – 3)2 _ (y + 1)2 hyperbola = 1 2) 1)