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

D AY 138 – E QUATION OF ELLIPSE

I NTRODUCTION 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 of distances from the foci is constant.

V OCABULARY Foci of an ellipse These are points on the main axis of the ellipse equidistant from the center and such that the sum of 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. 𝑧 𝐵 𝐺 2 𝐺 𝑦 1 𝐶 Points 𝐺 1 and 𝐺 2 are foci of the ellipse. 𝐺 1 𝐵 + 𝐵𝐺 2 = 𝐺 1 𝐶 + 𝐶𝐺 2 = 𝑑𝑝𝑜𝑡𝑢𝑏𝑜𝑢 We will use this fact to derive the equation of an ellipse.

Let the sum of distances from point A to the foci be 2𝑑 and the foci be at points (−𝑒, 0) and 𝑒, 0 . 𝑧 𝐵(𝑦, 𝑧) 𝐺 2 (−𝑒, 0) 𝐺 1 𝑒, 0 𝑦 𝐺 1 𝐵 + 𝐵𝐺 2 = 2𝑑 If we write this using the distance formula we get, 𝑦 − 𝑒 2 + 𝑧 2 + 𝑦 + 𝑒 2 + 𝑧 2 = 2𝑑 𝑦 − 𝑒 2 + 𝑧 2 = 2𝑑 − 𝑦 + 𝑒 2 + 𝑧 2 To eliminate the square root, we square both sides,

𝑦 − 𝑒 2 + 𝑧 2 = 4𝑑 2 − 4c 𝑦 + 𝑒 2 + 𝑧 2 + 𝑦 + 𝑒 2 + 𝑧 2 𝑦 2 − 2𝑒𝑦 + 𝑒 2 + 𝑧 2 = 4𝑑 2 − 4c 𝑦 + 𝑒 2 + 𝑧 2 + 𝑦 2 + 2𝑒𝑦 + 𝑒 2 + 𝑧 2 This simplifies to, 𝑦 + 𝑒 2 + 𝑧 2 = 4𝑑 2 + 4𝑒𝑦 dividing both sides by 4 we 4c 𝑦 + 𝑒 2 + 𝑧 2 = 𝑑 2 + 𝑒𝑦 get, c Squaring both sides we have, 𝑑 2 𝑦 2 + 2𝑒𝑦 + 𝑒 2 + 𝑧 2 = 𝑑 4 + 2𝑑 2 𝑒𝑦 + 𝑒 2 𝑦 2 𝑑 2 𝑦 2 + 𝑑 2 𝑒 2 𝑦 + 𝑑 2 𝑧 2 = 𝑑 4 + 2𝑑 2 𝑒𝑦 − 2𝑑 2 𝑒𝑦 + 𝑒 2 𝑦 2 𝑑 2 𝑦 2 − 𝑒 2 𝑦 2 + 𝑑 2 𝑧 2 = 𝑑 4 − 𝑑 2 𝑒 2 This simplifies to, 𝑑 2 − 𝑒 2 𝑦 2 + 𝑑 2 𝑧 2 = 𝑑 2 (𝑑 2 − 𝑒 2 )

𝑑 2 − 𝑒 2 𝑦 2 + 𝑑 2 𝑧 2 = 𝑑 2 𝑑 2 − 𝑒 2 … … (𝑗) In ∠𝐺 2 𝐵, 2𝑒 < 2𝑑 which implies that 𝑒 < 𝑑 and 1 𝐺 therefore 𝑑 2 − 𝑒 2 > 0. Let 𝑑 2 − 𝑒 2 = 𝑐 2 . equation (𝑗) becomes, 𝑐 2 𝑦 2 + 𝑑 2 𝑧 2 = 𝑑 2 𝑐 2 Dividing both sides by 𝑑 2 𝑐 2 we get, 𝑐 2 𝑦 2 𝑑 2 𝑧 2 𝑑 2 𝑐 2 𝑑 2 𝑐 2 + 𝑑 2 𝑐 2 = 𝑑 2 𝑐 2 𝒚 𝟑 𝒛 𝟑 𝒄 𝟑 = 𝟐 (this is the equation of the ellipse) 𝒅 𝟑 + Since 𝑑 2 − 𝑒 2 = 𝑐 2 then, 𝑑 2 − 𝑒 2 = 𝑐 2 , then 𝑐 < 𝑑 . To find the 𝑦 -intercepts, we substitute y with 0 in the equation.

𝑦 2 𝑑 2 = 1 ⟹ 𝑦 2 = 𝑑 2 The equation becomes 𝑦 2 = 𝑑 2 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 of the ellipse. Similarly the 𝑧 -intercepts are found by setting 𝑦 = 0, 𝒛 𝟑 The equation of the ellipse will become, 𝒄 𝟑 = 𝟐 , Thus 𝑧 2 = 𝑐 2 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.

. 𝑧 (0, 𝑐) (𝑑, 0) (−𝑑, 0) 𝐺 1 𝑒, 0 𝑦 𝐺 2 (−𝑒, 0) (0, −𝑐) 𝒚 𝟑 𝒛 𝟑 In this ellipse c > 𝑐, its equation is 𝒅 𝟑 + 𝒄 𝟑 = 𝟐 𝑒 2 = 𝑑 2 − 𝑐 2 The vertices are (±𝑑, 0) Note that are always the endpoints of the line segment joining the major axis of the ellipse.

𝑧 . (0, 𝑑) (0, 𝑒) (−𝑐, 0) (𝑐, 0) 0 𝑦 (0, −𝑒) (0, −𝑑) The major axis of this ellipse is the 𝑧 -axis. The vertices of the ellipse are (0, ±𝑑) 𝒚 𝟑 𝒛 𝟑 The equation of this ellipse is 𝒄 𝟑 + 𝒅 𝟑 = 𝟐 The has foci 0, ±𝑒 𝑒 2 = 𝑑 2 − 𝑐 2

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 (𝒚−𝒏) 𝟑 (𝒛−𝒐) 𝟑 this ellipse will be + = 𝟐 𝒅 𝟑 𝒄 𝟑

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. 𝒚 𝟑 𝒛 𝟑 The equation will have the notation 𝒅 𝟑 + 𝒄 𝟑 = 𝟐 Where 𝑒 2 = 𝑑 2 − 𝑐 2 and 𝑒 = 5, 𝑑 = 7 We get the value of b to find the equation. 5 2 = 7 2 − 𝑐 2 𝑐 2 = 49 − 25 = 24 24 , thus the 𝑧 -intercepts are ± 24, 0 𝑐 = 𝒚 𝟑 𝒛 𝟑 The equation of the ellipse is 𝟓𝟘 + 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. 𝒚 𝟑 𝒛 𝟑 The equation will have the notation 𝒄 𝟑 + 𝒅 𝟑 = 𝟐 𝑒 2 = 𝑑 2 − 𝑐 2 and 𝑑 = 6, 𝑒 = 3 3 2 = 6 2 − 𝑐 2 𝑐 2 = 36 − 9 = 27 𝒚 𝟑 𝒛 𝟑 The equation of the ellipse is 𝟑𝟖 + 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. 7+1 3+3 Its center will be at point , = (4,3) 2 2 7 − 4 2 + 3 − 3 2 Distance from center to vertex, c = = 3 Distance from the center the foci, 𝑒 = 6 − 4 2 + 3 − 3 2 = 2

𝑒 2 = 𝑑 2 − 𝑐 2 2 2 = 3 2 − 𝑐 2 𝑐 2 = 9 − 4 = 5 (𝒚−𝒏) 𝟑 (𝒛−𝒐) 𝟑 The notation of the equation is + = 𝟐 𝒅 𝟑 𝒄 𝟑 𝑛, 𝑜 = (4,3) (𝒚−𝟓) 𝟑 (𝒛−𝟒) 𝟑 The equation is + = 𝟐 𝟘 𝟔

Example 4 𝑦 2 𝑧 2 Given the equation 9 = 1 , find the vertices 25 + and the foci of this ellipse. Solution From this equation 𝑑 2 = 25 and 𝑐 2 = 9 Distance from the center to vertex 𝑑 = 5 Let the vertices be ±𝑦, 0 𝑦 − 0 2 + 0 − 0 2 = Thus 5 = 𝑦 𝑦 = 5 The vertices are ±5,0 Let d be distance from the center to one focus 𝑒 2 = 𝑑 2 − 𝑐 2 𝑒 2 = 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 .

A NSWERS TO HOMEWORK 𝒚 𝟑 𝒛 𝟑 𝟐𝟏𝟏 + 𝟒𝟕 = 𝟐

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