Just like the ellipse and the parabola, a hyperbola is formed when a - - PowerPoint PPT Presentation

DAY 139 – EQUATION OF A

HYPERBOLA

INTRODUCTION

In our prior conic sections lessons, we discussed in detail the two conic sections, the parabola, and the ellipse. The hyperbola is another conic section we have not yet explored. It also has foci and vertices just like the ellipses and parabolas. It is unique because it is separated into two parts called branches. The conic sections have a variety of applications in a number of fields ranging from engineering to space exploration. The shadows cast

• n a wall in a dark room by a lit lampshade, orbits

followed by comets about the sun are hyperbolic and cooling towers of nuclear reactors are hyperbolic in shape.

Just like the ellipse and the parabola, a hyperbola is formed when a plane intersects a double cone. In this lesson, we are going to derive the equation of a hyperbola from the foci using the fact that the difference in lengths of the two foci from the hyperbola is constant.

VOCABULARY

 Hyperbola

The set of all points in a plane for which the difference of distances from two fixed points to any point on the hyperbola to is constant.

 Asymptotes of a hyperbola

Straight lines which approach but neither cross nor touch the hyperbola and are not part of the real

• hyperbola. They intersect at the center of the

hyperbola.

A HYPERBOLA AS A RESULT OF

INTERSECTION OF A PLANE AND A CONE

 Hyperbola is the curve of intersection of a plane

parallel to the axis of the cone and the double cone.

Axis

A HYPERBOLA WITH CENTER AT THE ORIGIN

 A hyperbola whose center is at the origin can

either open towards the right and the left or upwards and downwards.

 The hyperbola on the next slide opens towards

the left and right.

 It consists of two vertices V

1 −𝑏, 0 and V2 𝑏, 0

 It has two foci F1 −𝑑, 0 and F2 𝑑, 0  It has two asymptotes, 𝑧 = ±

𝑐 𝑏.

 A hyperbola is split into two parts called

branches.

THE HYPERBOLA 𝒚𝟑

𝒃𝟑 − 𝒛𝟑 𝒄𝟑 = 𝟐

V2 𝑏, 0 F2 𝑑, 0 x y O F1 −𝑑, 0 V1 −𝑏, 0 𝑧 = − 𝑐 𝑏 𝑧 = 𝑐 𝑏

DERIVING THE EQUATION OF A PARABOLA

 We will use the geometric definition of a

hyperbola and the distance formula to derive the standard equations of a hyperbola on the coordinate plane.

 We will use a hyperbola placed in the coordinate

plane such that the foci are on the 𝑦 −axis and equidistant from the origin.

 The foci will be denoted by F1 and F2 and having

coordinates −𝑑, 0 and 𝑑, 0 respectively.

 According to the definition of a hyperbola, the

distance from the P to F1 is equal to the distance from P to F2.

𝐐𝐆𝟐 − 𝐐𝐆𝟑 = 𝐛 𝐝𝐩𝐨𝐭𝐮𝐛𝐨𝐮

F2 𝑑, 0 x y O F1 −𝑑, 0 𝑧 = − 𝑐 𝑏 𝑧 = 𝑐 𝑏 P 𝑦, 𝑧

 The center of the hyperbola is at the origin and

the hyperbola intersects the 𝑦 −axis at points V

1

and V2 having coordinates −𝑏, 0 and 𝑏, 0 respectively.

 The distance from either focus to the nearest

vertex is given by 𝑑 − 𝑏 while the distance from either focus to the furthest vertex is given by 𝑑 + 𝑏. The difference between this two distances is: 𝑑 + 𝑏 − 𝑑 − 𝑏 = 2𝑏 or 𝑑 − 𝑏 − 𝑑 + 𝑏 = −2𝑏

 This shows that ±2𝑏 is the common difference in

the distances between the foci and any point on the parabola.

 It then follows from the definition of a hyperbola

that: PF1 − PF2 = ±2𝑏

 Therefore by the distance formula, we have:

𝑦 + 𝑑 2 + 𝑧 − 0 2 − 𝑦 − 𝑑 2 + 𝑧 − 0 2 = ±2𝑏

 This simplifies to:

𝑦 + 𝑑 2 + 𝑧2 − 𝑦 − 𝑑 2 + 𝑧2 = ±2𝑏

 Adding on both

𝑦 − 𝑑 2 + 𝑧2 sides of the equation we have: 𝑦 + 𝑑 2 + 𝑧2 = ±2𝑏 + 𝑦 − 𝑑 2 + 𝑧2

 We square both sides of the equation to get rid of

the radicals. 𝑦 + 𝑑 2 + 𝑧2 = 4𝑏2 ± 4𝑏 𝑦 − 𝑑 2 + 𝑧2 + 𝑦 − 𝑑 2 + 𝑧2

 Subtracting 𝑦 − 𝑑 2 + 𝑧2 from both sides of the

equation, we have: 4𝑑𝑦 = 4𝑏2 ± 4𝑏 𝑦 − 𝑑 2 + 𝑧2

 Subtracting 4𝑏2 from both sides of the equation,

we have: 4𝑑𝑦 − 4𝑏2 = ±4𝑏 𝑦 − 𝑑 2 + 𝑧2

 Dividing both sides of the equation by 4, we have:

𝑑𝑦 − 𝑏2 = ±𝑏 𝑦 − 𝑑 2 + 𝑧2

 Squaring both sides of the equation again, we

have: 𝑑2𝑦2 − 2𝑏2𝑑𝑦 + 𝑏4 = 𝑏2 𝑦 − 𝑑 2 + 𝑧2

 Multiplying the right hand side, we have:

𝑑2𝑦2 − 2𝑏2𝑑𝑦 + 𝑏4 = 𝑏2𝑦2 − 2𝑏2𝑑𝑦 + 𝑏2𝑑2 + 𝑏2𝑧2

 Subtracting −2𝑏2𝑑𝑦 + 𝑏4 from both sides of the

equation, we have: 𝑑2𝑦2 = 𝑏2𝑦2 + 𝑏2𝑑2 + 𝑏2𝑧2 − 𝑏4

 Subtracting 𝑏2𝑑2 + 𝑏2𝑧2from both sides, we have:

𝑑2𝑦2 − 𝑏2𝑦2 − 𝑏2𝑧2 = 𝑏2𝑑2 − 𝑏4

 Factorizing both sides we have:

𝑦2 𝑑2 − 𝑏2 − 𝑏2𝑧2 = 𝑏2 𝑑2 − 𝑏2

 If we draw perpendiculars to the 𝑦 −axis at the

both vertices of the hyperbola as shown below, right triangles are formed.

 The right triangles will have one of their legs

along the 𝑦 −axis each of length 𝑏 and their hypotenuses will be each of length 𝑑. If the lengths of the other leg is 𝑐, then we can apply the Pythagorean theorem and have:

 𝑏2 + 𝑐2 = 𝑑2 which can also be written as:

𝑐2 = 𝑑2 − 𝑏2

 Now substituting 𝑐2 with 𝑑2 − 𝑏2in the equation

𝑦2 𝑑2 − 𝑏2 − 𝑏2𝑧2 = 𝑏2 𝑑2 − 𝑏2 , we have: 𝑦2𝑐2 − 𝑏2𝑧2 = 𝑏2𝑐2

 Dividing both sides of the equation by 𝑏2𝑐2, we

have: 𝑦2 𝑏2 − 𝑧2 𝑐2 = 1

 This is the equation a hyperbola with at the

• rigin.

𝑐 F2 𝑑, 0 x y O F1 −𝑑, 0 𝑧 = − 𝑐 𝑏 𝑧 = 𝑐 𝑏 P 𝑦, 𝑧 𝑏 𝑏 𝑐 𝑐 𝑑 𝑑 𝑐 𝑑 𝑑

THE HYPERBOLA 𝒚𝟑

𝒃𝟑 − 𝒛𝟑 𝒄𝟑 = 𝟐

 It has the foci on the 𝑦 −axis.  The foci are ±𝑑, 0 , where 𝑑2 = 𝑏2 + 𝑐2  The vertices are ±𝑏, 0  The asymptotes, the lines that contain the

hypotenuses of the right triangles have the equations:

 𝑧 = ±

𝑐 𝑏 𝑦

 It is also referred to as a horizontal hyperbola

because the axis of symmetry is x-axis.

 The center is at the origin.

THE HYPERBOLA 𝒛𝟑

𝒃𝟑 − 𝒚𝟑 𝒄𝟑 = 𝟐

 It has the foci on the 𝑧 −axis.  The foci are 0, ±𝑑 , where 𝑑2 = 𝑏2 + 𝑐2  The vertices are 0, ±𝑏  The asymptotes, the lines that contain the

hypotenuses of the right triangles have the equations:

 𝑧 = ±

𝑏 𝑐 𝑦

 It is also referred to as a vertical hyperbola

because the axis of symmetry is y-axis.

 The center is at the origin

THE HYPERBOLA WITH CENTER (ℎ, 𝑙)

 If the axis of symmetry is vertical, that is,

parallel to the y-axis, the hyperbola has the equation of the form: 𝒛 − 𝒍 𝟑 𝒃𝟑 − 𝒚 − 𝒊 𝟑 𝒄𝟑 = 𝟐

 If the axis of symmetry is horizontal, that is,

parallel to the x-axis, the hyperbola has the equation of the form: 𝒚 − 𝒊 𝟑 𝒃𝟑 − 𝒛 − 𝒍 𝟑 𝒄𝟑 = 𝟐

Example 1 Find the foci, the equation of the asymptotes and the intercepts of the hyperbola: 36𝑧2 − 25𝑦2 = 900 Solution We write the equation in the standard form fir a hyperbola by dividing both sides of the equation by

• 900. It becomes:

𝑧2 25 − 𝑦2 36 = 1 The first variable is 𝑧 hence it is a vertical hyperbola.

We compare it to the equation:

𝒛𝟑 𝒃𝟑 − 𝒚𝟑 𝒄𝟑 = 𝟐

The foci will be of the form 0, ±𝑑 where 𝑑2 = 𝑏2 + 𝑐2 Therefore 𝑑2 = 𝑏2 + 𝑐2 = 25 + 36 = 61 𝑑 = ± 61 The foci are 0, ± 61 : 𝟏, 𝟕𝟐 and 𝟏, − 𝟕𝟐 This is a vertical hyperbola, therefore it will have 𝑧 −intercepts. These will be its vertices 0, ±𝑏 𝑏2 = 25 ⇒ 𝑏 = ± 25 𝑏 = ±5 The 𝑧 −intercepts are 0, ±5 : 𝟏, 𝟔 and 𝟏, −𝟔

The asymptotes will be of the form: 𝑧 = ±

𝑏 𝑐 𝑦

𝑏 = 5 and 𝑐 = 6 The asymptotes are: 𝒛 =

𝟔 𝟕 𝒚 and 𝒛 = − 𝟔 𝟕 𝒚

Example 2 Find the equation of a hyperbola with foci ±10, 0 and vertices ±6, 0 Solution The center of the hyperbola is the origin because the foci are on the x-axis. The term in 𝑦2 is positive since foci lie on the x- axis. The equation will be of the form

𝒚𝟑 𝒃𝟑 − 𝒛𝟑 𝒄𝟑 = 𝟐

Using the vertices we have: 𝑏2 = 62 = 36 Using the foci we have: 𝑑2 = 102 = 100 We get the value of 𝑐 using the relation: 𝑑2 = 𝑏2 + 𝑐2 𝑐 = 100 − 36 = 8 The equation thus becomes:

𝒚𝟑 𝟒𝟕 − 𝒛𝟑 𝟕𝟓 = 𝟐

HOMEWORK Find the foci and the asymptotes of the hyperbola 2𝑧2 = 6𝑦2 + 24

ANSWERS TO HOMEWORK

The foci are 0, ±4 The asymptotes are the lines: 𝑧 = 3 𝑦 and 𝑧 = − 3 𝑦

THE END