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

D AY 139 – E QUATION OF A HYPERBOLA

I NTRODUCTION 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 on 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.

V OCABULARY  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 V 2 𝑏, 0  It has two foci F 1 −𝑑, 0 and F 2 𝑑, 0 𝑐  It has two asymptotes, 𝑧 = ± 𝑏 .  A hyperbola is split into two parts called branches.

T HE HYPERBOLA 𝒚 𝟑 𝒃 𝟑 − 𝒛 𝟑 𝒄 𝟑 = 𝟐 y 𝑧 = 𝑐 𝑏 𝑧 = − 𝑐 𝑏 F 2 𝑑, 0 F 1 −𝑑, 0 x V 1 −𝑏, 0 V 2 𝑏, 0 O

D ERIVING 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 F 1 and F 2 and having coordinates −𝑑, 0 and 𝑑, 0 respectively.  According to the definition of a hyperbola, the distance from the P to F 1 is equal to the distance from P to F 2 .

𝐐𝐆 𝟐 − 𝐐𝐆 𝟑 = 𝐛 𝐝𝐩𝐨𝐭𝐮𝐛𝐨𝐮 y 𝑧 = 𝑐 𝑧 = − 𝑐 𝑏 𝑏 P 𝑦, 𝑧 O x F 1 −𝑑, 0 F 2 𝑑, 0

 The center of the hyperbola is at the origin and the hyperbola intersects the 𝑦 − axis at points V 1 and V 2 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: PF 1 − PF 2 = ±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 + 𝑧 2 − 𝑦 − 𝑑 2 + 𝑧 2 = ±2𝑏 𝑦 − 𝑑 2 + 𝑧 2 sides of the  Adding on both equation we have: 𝑦 + 𝑑 2 + 𝑧 2 = ±2𝑏 + 𝑦 − 𝑑 2 + 𝑧 2

 We square both sides of the equation to get rid of the radicals. 𝑦 + 𝑑 2 + 𝑧 2 = 4𝑏 2 ± 4𝑏 𝑦 − 𝑑 2 + 𝑧 2 + 𝑦 − 𝑑 2 + 𝑧 2  Subtracting 𝑦 − 𝑑 2 + 𝑧 2 from both sides of the equation, we have: 4𝑑𝑦 = 4𝑏 2 ± 4𝑏 𝑦 − 𝑑 2 + 𝑧 2  Subtracting 4𝑏 2 from both sides of the equation, we have: 4𝑑𝑦 − 4𝑏 2 = ±4𝑏 𝑦 − 𝑑 2 + 𝑧 2

 Dividing both sides of the equation by 4, we have: 𝑑𝑦 − 𝑏 2 = ±𝑏 𝑦 − 𝑑 2 + 𝑧 2  Squaring both sides of the equation again, we have: 𝑑 2 𝑦 2 − 2𝑏 2 𝑑𝑦 + 𝑏 4 = 𝑏 2 𝑦 − 𝑑 2 + 𝑧 2  Multiplying the right hand side, we have: 𝑑 2 𝑦 2 − 2𝑏 2 𝑑𝑦 + 𝑏 4 = 𝑏 2 𝑦 2 − 2𝑏 2 𝑑𝑦 + 𝑏 2 𝑑 2 + 𝑏 2 𝑧 2  Subtracting −2𝑏 2 𝑑𝑦 + 𝑏 4 from both sides of the equation, we have: 𝑑 2 𝑦 2 = 𝑏 2 𝑦 2 + 𝑏 2 𝑑 2 + 𝑏 2 𝑧 2 − 𝑏 4

 Subtracting 𝑏 2 𝑑 2 + 𝑏 2 𝑧 2 from both sides, we have: 𝑑 2 𝑦 2 − 𝑏 2 𝑦 2 − 𝑏 2 𝑧 2 = 𝑏 2 𝑑 2 − 𝑏 4  Factorizing both sides we have: 𝑦 2 𝑑 2 − 𝑏 2 − 𝑏 2 𝑧 2 = 𝑏 2 𝑑 2 − 𝑏 2  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:

 𝑏 2 + 𝑐 2 = 𝑑 2 which can also be written as: 𝑐 2 = 𝑑 2 − 𝑏 2  Now substituting 𝑐 2 with 𝑑 2 − 𝑏 2 in the equation 𝑦 2 𝑑 2 − 𝑏 2 − 𝑏 2 𝑧 2 = 𝑏 2 𝑑 2 − 𝑏 2 , we have: 𝑦 2 𝑐 2 − 𝑏 2 𝑧 2 = 𝑏 2 𝑐 2  Dividing both sides of the equation by 𝑏 2 𝑐 2 , we have: 𝑦 2 𝑏 2 − 𝑧 2 𝑐 2 = 1  This is the equation a hyperbola with at the origin.

y 𝑧 = 𝑐 𝑧 = − 𝑐 𝑏 𝑏 P 𝑦, 𝑧 𝑑 𝑑 𝑐 𝑐 x O 𝑏 𝑏 F 1 −𝑑, 0 F 2 𝑑, 0 𝑐 𝑐 𝑑 𝑑

T HE HYPERBOLA 𝒚 𝟑 𝒃 𝟑 − 𝒛 𝟑 𝒄 𝟑 = 𝟐  It has the foci on the 𝑦 − axis.  The foci are ±𝑑, 0 , where 𝑑 2 = 𝑏 2 + 𝑐 2  The vertices are ±𝑏, 0  The asymptotes, the lines that contain the hypotenuses of the right triangles have the equations: 𝑐  𝑧 = ± 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.

T HE HYPERBOLA 𝒛 𝟑 𝒃 𝟑 − 𝒚 𝟑 𝒄 𝟑 = 𝟐  It has the foci on the 𝑧 − axis.  The foci are 0, ±𝑑 , where 𝑑 2 = 𝑏 2 + 𝑐 2  The vertices are 0, ±𝑏  The asymptotes, the lines that contain the hypotenuses of the right triangles have the equations: 𝑏  𝑧 = ± 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

T HE 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 − 𝒚 − 𝒊 𝟑 = 𝟐 𝒃 𝟑 𝒄 𝟑  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 − 𝒛 − 𝒍 𝟑 = 𝟐 𝒃 𝟑 𝒄 𝟑

Example 1 Find the foci, the equation of the asymptotes and the intercepts of the hyperbola: 36𝑧 2 − 25𝑦 2 = 900 Solution We write the equation in the standard form fir a hyperbola by dividing both sides of the equation by 900. It becomes: 𝑧 2 25 − 𝑦 2 36 = 1 The first variable is 𝑧 hence it is a vertical hyperbola.

𝒛 𝟑 𝒚 𝟑 We compare it to the equation: 𝒃 𝟑 − 𝒄 𝟑 = 𝟐 The foci will be of the form 0, ±𝑑 where 𝑑 2 = 𝑏 2 + 𝑐 2 Therefore 𝑑 2 = 𝑏 2 + 𝑐 2 = 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 𝑏 2 = 25 ⇒ 𝑏 = ± 25 𝑏 = ±5 The 𝑧 − intercepts are 0, ±5 : 𝟏, 𝟔 and 𝟏, −𝟔

𝑏 The asymptotes will be of the form: 𝑧 = ± 𝑐 𝑦 𝑏 = 5 and 𝑐 = 6 𝟔 𝟔 The asymptotes are: 𝒛 = 𝟕 𝒚 and 𝒛 = − 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 𝑦 2 is positive since foci lie on the x- axis. 𝒚 𝟑 𝒛 𝟑 The equation will be of the form 𝒃 𝟑 − 𝒄 𝟑 = 𝟐

Using the vertices we have: 𝑏 2 = 6 2 = 36 Using the foci we have: 𝑑 2 = 10 2 = 100 We get the value of 𝑐 using the relation: 𝑑 2 = 𝑏 2 + 𝑐 2 𝑐 = 100 − 36 = 8 𝒚 𝟑 𝒛 𝟑 The equation thus becomes: 𝟒𝟕 − 𝟕𝟓 = 𝟐

HOMEWORK Find the foci and the asymptotes of the hyperbola 2𝑧 2 = 6𝑦 2 + 24

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

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