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6-3 Conic Sections: Ellipses

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1 6-3 Conic Sections: Ellipses
Geometric Definition: The intersection of a cone and a plane such that the plane is oblique to the base of the cone. (A circle is a special case of an ellipse where the plane is parallel to the base of the cone.) Algebraic definition: The set of all points in the plane such that the sum of the distances from two fixed points, called foci, remains constant.

2 So, about those foci . . . foci
From each point in the plane, the sum of the distances to the foci is a constant. Example: B A d1 d2 d1 d2 x f1 f2 Point A: d1+d2 = c foci Point B: d1+d2 = c y

3 Ellipse Terminology ๐‘ฅ 2 ๐‘Ž 2 + ๐‘ฆ 2 ๐‘ 2 =1 ๐‘ฆ 2 ๐‘Ž 2 + ๐‘ฅ 2 ๐‘ 2 =1
(0,a) 2๐‘Ž =๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘š๐‘Ž๐‘—๐‘œ๐‘Ÿ ๐‘Ž๐‘ฅ๐‘–๐‘  Major axis 2๐‘ =๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘š๐‘–๐‘›๐‘œ๐‘Ÿ๐‘Ž๐‘ฅ๐‘–๐‘  (0,b) Minor axis f1 (0,c) Major axis (๏€ญc,0) x Center (c,0) x Minor axis vertex (๏€ญa,0) (a,0) (๏€ญb,0) (b,0) f2 foci f1 Center vertex foci f2 (0,๏€ญ b) (0,๏€ญc) ๐‘ 2 = ๐‘Ž 2 โˆ’ ๐‘ 2 โ„Ž, ๐‘˜ =๐‘๐‘’๐‘›๐‘ก๐‘’๐‘Ÿ y (0,๏€ญ a) ๐‘“ 1 , ๐‘“ 2 =๐‘“๐‘œ๐‘๐‘– y ๐‘ฅ 2 ๐‘Ž ๐‘ฆ 2 ๐‘ 2 =1 ๐‘ฆ 2 ๐‘Ž ๐‘ฅ 2 ๐‘ 2 =1 Discuss definitions of center, foci, major axis (longer one) and minor axis (shorter one). Stress that the focis are always on the major axis. ๐‘ฅโˆ’โ„Ž 2 ๐‘Ž ๐‘ฆโˆ’๐‘˜ 2 ๐‘ 2 =1 ๐‘ฆโˆ’๐‘˜ 2 ๐‘Ž ๐‘ฅโˆ’โ„Ž 2 ๐‘ 2 =1 ๐‘“๐‘œ๐‘๐‘– ๐‘Ž๐‘Ÿ๐‘’ ๐‘Ž๐‘™๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘œ๐‘› ๐‘š๐‘Ž๐‘—๐‘œ๐‘Ÿ ๐‘Ž๐‘ฅ๐‘–๐‘  (๐‘™๐‘œ๐‘›๐‘”๐‘’๐‘Ÿ ๐‘œ๐‘›๐‘’) ๐‘Ž, ๐‘ ๐‘Ž๐‘Ÿ๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’๐‘  ๐‘Ž๐‘ค๐‘Ž๐‘ฆ ๐‘“๐‘Ÿ๐‘œ๐‘š โ„Ž,๐‘˜ ๐‘™๐‘Ž๐‘Ÿ๐‘”๐‘’๐‘Ÿ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘–๐‘› ๐‘‘๐‘’๐‘›๐‘œ๐‘š๐‘–๐‘›๐‘Ž๐‘ก๐‘œ๐‘Ÿ ๐‘–๐‘  ๐‘Ž 2

4 Example 1: ๐‘ฅ ๐‘ฆ 2 9 =1 Step 1: Identify if the ellipse is horizontal (a2 with x-term) or vertical (a2 with y-term). This ellipse is horizontal since 16 larger term and with x-term. Step 2: Identify and plot the center (h,k). This ellipse has a center at (0,0). Step 3: Plot the endpoints of the major axis. The major axis is horizontal so plot |a| units left and right from center. Since a2=16, a=4; therefore plot 4 units left and right of center. Step 4: Plot the endpoints of the minor axis. The minor axis is vertical so plot |b| units above and below the center. Since b2=9, b=3; therefore plot 3 units above and below center. Step 5: Calculate and plot foci. b2=a2 - c2 9 = 16 - c2; c2=7; c = ๏‚ฑ Since foci are on major axis (horizontal in this case), plot 2.65 units left and right of center. Step 6: Connect endpoint of axes with smooth curve.

5 Example 2: ๐‘ฅ ๐‘ฆ =1 Step 1: Identify if the ellipse is horizontal (a2 with x-term) or vertical (a2 with y-term). This ellipse is vertical since 81 larger term and with y-term. Step 2: Identify and plot the center (h,k). This ellipse has a center at (0,0). Step 3: Plot the endpoints of the major axis. The major axis is vertical so plot |a| units above and below center. Since a2=81, a=9; therefore plot 9 units above and below center. Step 4: Plot the endpoints of the minor axis. The minor axis is horizontal so plot |b| units left and right from the center. Since b2=36; b=6; therefore plot 6 units left and right of center. Step 5: Calculate and plot foci. b2=a2 - c2 36= 81 - c2; c2=45; c = ๏‚ฑ6.7. Since foci are on major axis (vertical in this case), plot 6.7 units above and below the center. Step 6: Connect endpoint of axes with smooth curve.

6 Example 3: (๐‘ฅ+5) (๐‘ฆโˆ’4) =1 Step 1: Identify if the ellipse is horizontal (a2 with x-term) or vertical (a2 with y-term). This ellipse is horizontal since 25 larger term and with x-term. Step 2: Identify and plot the center (h,k). This ellipse has a center at (๏€ญ5,4). Step 3: Plot the endpoints of the major axis. The major axis is horizontal so plot |a| units left and right from center. Since a2=25, a=5; therefore plot 5 units left and right of center. Step 4: Plot the endpoints of the minor axis. The minor axis is vertical so plot |b| units above and below the center. Since b2=16, b=4; therefore plot 4 units above and below center. Step 5: Calculate and plot foci. b2=a2 - c = 25 - c2; c2=9; c = ๏‚ฑ3. Since foci are on major axis (horizontal in this case), plot 3 units left and right of center. Step 6: Connect endpoint of axes with smooth curve.

7 How to enter into a calculator.
(๐‘ฅ+5) (๐‘ฆโˆ’4) =1 16(๐‘ฅ+5) (๐‘ฆโˆ’4) 2 =16 Multiply by 16 to isolate y-term. (๐‘ฆโˆ’4) 2 =16โˆ’ 16(๐‘ฅ+5) 2 25 Subtract x-term from both sides. ๐‘ฆโˆ’4=ยฑ 16โˆ’ 16(๐‘ฅ+5) 2 25 Take the square root of both sides. ๐‘ฆ=4ยฑ 16โˆ’ 16(๐‘ฅ+5) 2 25 Add 4 to both sides. ๐‘ฆ=4+{1,โˆ’1}โˆ— 16โˆ’ (๐‘ฅ+5) 2 Enter in y-editor of calculator.

8 6-4 Conic Sections: Hyperbolas
Geometric Definition: The intersection of a cone and a plane such that the plane is perpendicular to the base of the cone. Algebraic definition: The set of all points in the plane such that the difference of the distances from two fixed points, called foci, remains constant.

9 Hyperbola Terminology
2๐‘Ž =๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘ก๐‘Ÿ๐‘Ž๐‘›๐‘ ๐‘ฃ๐‘’๐‘Ÿ๐‘ ๐‘’ ๐‘Ž๐‘ฅ๐‘–๐‘  2๐‘ =๐‘™๐‘’๐‘›๐‘”๐‘กโ„Ž ๐‘๐‘œ๐‘›๐‘—๐‘ข๐‘”๐‘Ž๐‘ก๐‘’ ๐‘Ž๐‘ฅ๐‘–๐‘  (0,c) f1 Conjugate axis vertex (0,a) Transverse axis (0,b) (๏€ญc,0) x (c,0) Center f1 x Center Conjugate axis f2 (๏€ญa,0) (a,0) foci (๏€ญb,0) (b,0) vertex (0,๏€ญ b) (0,๏€ญ a) Transverse axis ๐‘ 2 = ๐‘ 2 โˆ’ ๐‘Ž 2 f2 foci (0,๏€ญc) โ„Ž, ๐‘˜ =๐‘๐‘’๐‘›๐‘ก๐‘’๐‘Ÿ y ๏‚ฑ ๐‘ ๐‘Ž slope of asymptotes ๐‘“ 1 , ๐‘“ 2 =๐‘“๐‘œ๐‘๐‘– y ๏‚ฑ ๐‘Ž ๐‘ slope of asymptotes ๐‘ฅ 2 ๐‘Ž 2 โˆ’ ๐‘ฆ 2 ๐‘ 2 =1 ๐‘ฆ 2 ๐‘Ž 2 โˆ’ ๐‘ฅ 2 ๐‘ 2 =1 ๐‘ฆโˆ’๐‘˜ 2 ๐‘Ž 2 โˆ’ ๐‘ฅโˆ’โ„Ž 2 ๐‘ 2 =1 ๐‘ฅโˆ’โ„Ž 2 ๐‘Ž 2 โˆ’ ๐‘ฆโˆ’๐‘˜ 2 ๐‘ 2 =1 ๐‘“๐‘œ๐‘๐‘– ๐‘Ž๐‘Ÿ๐‘’ ๐‘Ž๐‘™๐‘ค๐‘Ž๐‘ฆ๐‘  ๐‘œ๐‘› ๐‘ก๐‘Ÿ๐‘Ž๐‘›๐‘ ๐‘ฃ๐‘’๐‘Ÿ๐‘ ๐‘’ ๐‘Ž๐‘ฅ๐‘–๐‘  (๐‘๐‘œ๐‘›๐‘ก๐‘Ž๐‘–๐‘›๐‘  ๐‘ฃ๐‘’๐‘Ÿ๐‘ก๐‘–๐‘๐‘’๐‘ ) ๐‘Ž, ๐‘ ๐‘Ž๐‘Ÿ๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ž๐‘›๐‘๐‘’๐‘  ๐‘Ž๐‘ค๐‘Ž๐‘ฆ ๐‘“๐‘Ÿ๐‘œ๐‘š โ„Ž,๐‘˜ ๐‘Ž 2 ๐‘–๐‘  ๐‘–๐‘› ๐‘“๐‘–๐‘Ÿ๐‘ ๐‘ก ๐‘ก๐‘’๐‘Ÿ๐‘š ๐‘๐‘’๐‘“๐‘œ๐‘Ÿ๐‘’ ๐‘ ๐‘ข๐‘๐‘ก๐‘Ÿ๐‘Ž๐‘๐‘ก๐‘–๐‘œ๐‘› ๐‘ ๐‘–๐‘”๐‘›

10 Example 1: ๐‘ฅ โˆ’ ๐‘ฆ 2 4 =1 Step 1: Identify if the hyperbola is horizontal (x-term first) or vertical (y-term first). This hyperbola is horizontal since x-term appears first. Step 2: Identify and plot the center (h,k). This hyperbola has a center at (0,0). Step 3: Plot the endpoints of the transverse axis. The transverse axis is horizontal so plot |a| units left and right from center. Since a2=49, a=7; therefore plot 7 units left and right of center. Step 4: Plot the endpoints of the conjugate axis. The conjugate axis is vertical so plot |b| units above and below the center. Since b2=4, b=2; therefore plot 2 units above and below center. Step 5: Draw an a x b rectangle such that each of the axes endpoints is the midpoint of a side of the rectangle; draw the diagonals and extend them. The diagonals are the asymptotes Step 6: Sketch each branch of the hyperbola so that it approaches the asymptotes and passes through the vertex. Step 7: Calculate and plot foci. b2=c2 - a2 4 = c2-49; c2=53; c = ๏‚ฑ7.3. Since foci are on transverse axis (horizontal in this case), plot 7.3 units left and right of center.

11 ๐‘ฆ โˆ’ ๐‘ฅ =1 Example 2: Step 1: Identify if the hyperbola is horizontal (x-term first) or vertical (y-term first). This hyperbola is vertical since y-term appears first. Step 2: Identify and plot the center (h,k). This hyperbola has a center at (0,0). Step 3: Plot the endpoints of the transverse axis. The transverse axis is vertical so plot |a| units above and below the center. Since a2=36, a=6; therefore plot 6 units above and below the center. Step 4: Plot the endpoints of the conjugate axis. The conjugate axis is horizontal so plot |b| units left and right of the center. Since b2=64, b=8; therefore plot 8 units left and right of the center. Step 5: Draw an a x b rectangle such that each of the axes endpoints is the midpoint of a side of the rectangle; draw the diagonals and extend them. The diagonals are the asymptotes Step 6: Sketch each branch of the hyperbola so that it approaches the asymptotes and passes through the vertex. Step 7: Calculate and plot foci. b2=c2 - a = c2-36; c2=100; c = ๏‚ฑ10. Since foci are on transverse axis (vertical in this case), plot 10 units above and below the center.

12 Example 3: (๐‘ฅโˆ’4) โˆ’ (๐‘ฆโˆ’6) =1 Step 1: Identify if the hyperbola is horizontal (x-term first) or vertical (y-term first). This hyperbola is horizontal since x-term appears first. Step 2: Identify and plot the center (h,k). This hyperbola has a center at (4,6). Step 3: Plot the endpoints of the transverse axis. The transverse axis is horizontal so plot |a| units left and right from center. Since a2=25, a=5; therefore plot 5 units left and right of center. Step 4: Plot the endpoints of the conjugate axis. The conjugate axis is vertical so plot |b| units above and below the center. Since b2=36, b=6; therefore plot 6 units above and below center. Step 5: Draw an a x b rectangle such that each of the axes endpoints is the midpoint of a side of the rectangle; draw the diagonals and extend them. The diagonals are the asymptotes Step 6: Sketch each branch of the hyperbola so that it approaches the asymptotes and passes through the vertex. Step 7: Calculate and plot foci. b2=c2 - a = c2-25; c2=61; c = ๏‚ฑ7.8. Since foci are on transverse axis (horizontal in this case), plot 7.8 units left and right of center.

13 How to enter into a calculator.
(๐‘ฅโˆ’4) โˆ’ (๐‘ฆโˆ’6) =1 36(๐‘ฅโˆ’4) โˆ’ (๐‘ฆโˆ’6) 2 =36 Multiply by 36 to isolate y-term. 36(๐‘ฅโˆ’4) โˆ’ 36 = (๐‘ฆโˆ’6) 2 Subtract constant from both sides and add y-term to both sides ๏‚ฑ 36(๐‘ฅโˆ’4) โˆ’ 36 =๐‘ฆโˆ’6 Take the square root of both sides. 6๏‚ฑ 36(๐‘ฅโˆ’4) โˆ’ 36 =๐‘ฆ Add 6 to both sides. ๐‘ฆ=6= 1,โˆ’1 โˆ— (๐‘ฅโˆ’4) 2 โˆ’ 36 Enter in y-editor of calculator.


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