If a plane cuts through a cone such that the plane is not parallel - - PowerPoint PPT Presentation

DAY 138 – EQUATION OF

ELLIPSE

INTRODUCTION

A plane cutting through a circular cone produces different shapes depending on the angle the plane makes with the circular cone. If the plane cuts across the cone such that it is not perpendicular to the vertical height of the cone, a symmetrical figure in a shape of a stretched circle is produced. This figure is referred to as the ellipse. In this lesson we will derive the equations of ellipses given the foci, using the fact that the sum

• f distances from the foci is constant.

VOCABULARY

Foci of an ellipse These are points on the main axis of the ellipse equidistant from the center and such that the sum

• f distances from a point to the foci is constant.

Ellipse These are set of points in a plane with two foci such that the sum of distances from the foci to each point in the set is constant.

If a plane cuts through a cone such that the plane is not parallel to the sliding side of the cone and not perpendicular to its vertical height, their intersection produces a curved shape . This shape is the ellipse. The dotted line forms the shape of an ellipse on the cutting plane.

A set of points in a plane with the sum of distances from foci being constant is called an ellipse. An ellipse has two foci. Consider the figure below. Points 𝐺

1 and 𝐺 2 are foci of the ellipse.

𝐺

1𝐵 + 𝐵𝐺 2 = 𝐺 1𝐶 + 𝐶𝐺 2 = 𝑑𝑝𝑜𝑡𝑢𝑏𝑜𝑢

We will use this fact to derive the equation of an ellipse.

𝐺2 𝐵 𝐺

1

𝐶 𝑧 𝑦

Let the sum of distances from point A to the foci be 2𝑑 and the foci be at points (−𝑒, 0) and 𝑒, 0 . 𝐺

1𝐵 + 𝐵𝐺 2 = 2𝑑

If we write this using the distance formula we get, 𝑦 − 𝑒 2 + 𝑧2 + 𝑦 + 𝑒 2 + 𝑧2 = 2𝑑 𝑦 − 𝑒 2 + 𝑧2 = 2𝑑 − 𝑦 + 𝑒 2 + 𝑧2 To eliminate the square root, we square both sides,

𝐺2(−𝑒, 0) 𝐵(𝑦, 𝑧) 𝐺

1 𝑒, 0

𝑧 𝑦

𝑦 − 𝑒 2 + 𝑧2 = 4𝑑2 − 4c 𝑦 + 𝑒 2 + 𝑧2 + 𝑦 + 𝑒 2 + 𝑧2 𝑦2 − 2𝑒𝑦 + 𝑒2 + 𝑧2 = 4𝑑2 − 4c 𝑦 + 𝑒 2 + 𝑧2 + 𝑦2 + 2𝑒𝑦 + 𝑒2 + 𝑧2 This simplifies to, 4c 𝑦 + 𝑒 2 + 𝑧2 = 4𝑑2 + 4𝑒𝑦 dividing both sides by 4 we get, c 𝑦 + 𝑒 2 + 𝑧2 = 𝑑2 + 𝑒𝑦 Squaring both sides we have, 𝑑2 𝑦2 + 2𝑒𝑦 + 𝑒2 + 𝑧2 = 𝑑4 + 2𝑑2𝑒𝑦 + 𝑒2𝑦2 𝑑2𝑦2 + 𝑑2𝑒2𝑦 + 𝑑2𝑧2 = 𝑑4 + 2𝑑2𝑒𝑦 − 2𝑑2𝑒𝑦 + 𝑒2𝑦2 𝑑2𝑦2 − 𝑒2𝑦2 + 𝑑2𝑧2 = 𝑑4 − 𝑑2𝑒2 This simplifies to, 𝑑2 − 𝑒2 𝑦2 + 𝑑2𝑧2 = 𝑑2(𝑑2 − 𝑒2)

𝑑2 − 𝑒2 𝑦2 + 𝑑2𝑧2 = 𝑑2 𝑑2 − 𝑒2 … … (𝑗) In ∠𝐺

1𝐺 2𝐵, 2𝑒 < 2𝑑 which implies that 𝑒 < 𝑑 and

therefore 𝑑2 − 𝑒2 > 0. Let 𝑑2 − 𝑒2 = 𝑐2. equation (𝑗) becomes, 𝑐2𝑦2 + 𝑑2𝑧2 = 𝑑2𝑐2 Dividing both sides by 𝑑2𝑐2 we get,

𝑐2𝑦2 𝑑2𝑐2 + 𝑑2𝑧2 𝑑2𝑐2 = 𝑑2𝑐2 𝑑2𝑐2 𝒚𝟑 𝒅𝟑 + 𝒛𝟑 𝒄𝟑 = 𝟐 (this is the equation of the ellipse)

Since 𝑑2 − 𝑒2 = 𝑐2 then, 𝑑2 − 𝑒2 = 𝑐2, then 𝑐 < 𝑑. To find the 𝑦-intercepts, we substitute y with 0 in the equation.

The equation becomes

𝑦2 𝑑2 = 1 ⟹ 𝑦2 = 𝑑2

𝑦2 = 𝑑2 implies that 𝑦 = ±𝑑 therefore the coordinates of 𝑦-intercepts are (−𝑑, 0) and (𝑑, 0). The coordinates (−𝑑, 0) and (𝑑, 0) are called the vertices

• f the ellipse.

Similarly the 𝑧-intercepts are found by setting 𝑦 = 0, The equation of the ellipse will become,

𝒛𝟑 𝒄𝟑 = 𝟐,

Thus 𝑧2 = 𝑐2 implies that y = ±𝑐 therefore the coordinates of 𝑧-intercepts are (−𝑐, 0) and (𝑐, 0). The ellipse is symmetric on both x and y axes.

. In this ellipse c> 𝑐, its equation is

𝒚𝟑 𝒅𝟑 + 𝒛𝟑 𝒄𝟑 = 𝟐

𝑒2 = 𝑑2 − 𝑐2 The vertices are (±𝑑, 0) Note that are always the endpoints of the line segment joining the major axis of the ellipse.

𝐺2(−𝑒, 0) 𝐺

1 𝑒, 0

𝑧 𝑦 (𝑑, 0) (−𝑑, 0) (0, 𝑐) (0, −𝑐)

. The major axis of this ellipse is the 𝑧-axis. The vertices of the ellipse are (0, ±𝑑) The equation of this ellipse is

𝒚𝟑 𝒄𝟑 + 𝒛𝟑 𝒅𝟑 = 𝟐

The has foci 0, ±𝑒 𝑒2 = 𝑑2 − 𝑐2

(0, 𝑑) (0, −𝑑) (−𝑐, 0) (𝑐, 0) 𝑦 𝑧 (0, −𝑒) (0, 𝑒)

At times we may have an ellipse with is not centered at the origin. In this case the ellipse has been translated from the origin through a certain distance. If the ellipse is centered at point 𝑛, 𝑜 the equation of this ellipse will be

(𝒚−𝒏)𝟑 𝒅𝟑

+

(𝒛−𝒐)𝟑 𝒄𝟑

= 𝟐

Example 1 Find the equation of the ellipse with foci (±5,0) and vertices ±7,0 . Solution Since the foci and the vertices lie on the 𝑦-axis the major axis of the ellipse will be along 𝑦-axis. The equation will have the notation

𝒚𝟑 𝒅𝟑 + 𝒛𝟑 𝒄𝟑 = 𝟐

Where 𝑒2 = 𝑑2 − 𝑐2 and 𝑒 = 5, 𝑑 = 7 We get the value of b to find the equation. 52 = 72 − 𝑐2 𝑐2 = 49 − 25 = 24 𝑐 = 24 , thus the 𝑧-intercepts are ± 24, 0 The equation of the ellipse is

𝒚𝟑 𝟓𝟘 + 𝒛𝟑 𝟑𝟓 = 𝟐

Example 2 Find the equation of the ellipse with foci (0, ±3) and vertices 0, ±6 . Solution Since the foci and the vertices lie on the 𝑧-axis the major axis of the ellipse will be along the 𝑧-axis. The equation will have the notation

𝒚𝟑 𝒄𝟑 + 𝒛𝟑 𝒅𝟑 = 𝟐

𝑒2 = 𝑑2 − 𝑐2 and 𝑑 = 6, 𝑒 = 3 32 = 62 − 𝑐2 𝑐2 = 36 − 9 =27 The equation of the ellipse is

𝒚𝟑 𝟑𝟖 + 𝒛𝟑 𝟒𝟕 = 𝟐

Example 3 Find the equation of the ellipse with one focus at (6,3) and vertices 7,3 and (1,3) Solution The vertices have the 𝑧 coordinate as 3 which implies that the major axis of the ellipse is along the line 𝑧 = 3. Obviously the ellipse will be centered half way between the vertices. Its center will be at point

7+1 2

,

3+3 2

= (4,3) Distance from center to vertex, c = 7 − 4 2 + 3 − 3 2 = 3 Distance from the center the foci, 𝑒 = 6 − 4 2 + 3 − 3 2 = 2

𝑒2 = 𝑑2 − 𝑐2 22 = 32 − 𝑐2 𝑐2 = 9 − 4 = 5 The notation of the equation is

(𝒚−𝒏)𝟑 𝒅𝟑

+

(𝒛−𝒐)𝟑 𝒄𝟑

= 𝟐 𝑛, 𝑜 = (4,3) The equation is

(𝒚−𝟓)𝟑 𝟘

+

(𝒛−𝟒)𝟑 𝟔

= 𝟐

Example 4 Given the equation

𝑦2 25 + 𝑧2 9 = 1, find the vertices

and the foci of this ellipse. Solution From this equation 𝑑2 = 25 and 𝑐2 = 9 Distance from the center to vertex 𝑑 = 5 Let the vertices be ±𝑦, 0 Thus 5= 𝑦 − 0 2 + 0 − 0 2 = 𝑦 𝑦 = 5 The vertices are ±5,0 Let d be distance from the center to one focus 𝑒2 = 𝑑2 − 𝑐2 𝑒2 = 25 − 9 , 𝑒 = 4 The foci are at points ±4,0

HOMEWORK Find the equation of the ellipse with foci (±8,0) and vertices ±10,0 .

ANSWERS TO HOMEWORK

𝒚𝟑 𝟐𝟏𝟏 + 𝒛𝟑 𝟒𝟕 = 𝟐

THE END