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.

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

Ellipse Terminology 𝑥 2 𝑎 2 + 𝑦 2 𝑏 2 =1 𝑦 2 𝑎 2 + 𝑥 2 𝑏 2 =1 (0,a) 2𝑎 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠 Major axis 2𝑏 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑚𝑖𝑛𝑜𝑟𝑎𝑥𝑖𝑠 (0,b) Minor axis f1 (0,c) Major axis (c,0) x Center (c,0) x Minor axis vertex (a,0) (a,0) (b,0) (b,0) f2 foci f1 Center vertex foci f2 (0, b) (0,c) 𝑏 2 = 𝑎 2 − 𝑐 2 ℎ, 𝑘 =𝑐𝑒𝑛𝑡𝑒𝑟 y (0, a) 𝑓 1 , 𝑓 2 =𝑓𝑜𝑐𝑖 y 𝑥 2 𝑎 2 + 𝑦 2 𝑏 2 =1 𝑦 2 𝑎 2 + 𝑥 2 𝑏 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. 𝑥−ℎ 2 𝑎 2 + 𝑦−𝑘 2 𝑏 2 =1 𝑦−𝑘 2 𝑎 2 + 𝑥−ℎ 2 𝑏 2 =1 𝑓𝑜𝑐𝑖 𝑎𝑟𝑒 𝑎𝑙𝑤𝑎𝑦𝑠 𝑜𝑛 𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠 (𝑙𝑜𝑛𝑔𝑒𝑟 𝑜𝑛𝑒) 𝑎, 𝑏 𝑎𝑟𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑠 𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 ℎ,𝑘 𝑙𝑎𝑟𝑔𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑖𝑠 𝑎 2

Example 1: 𝑥 2 16 + 𝑦 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 = 2.65. 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.

Example 2: 𝑥 2 36 + 𝑦 2 81 =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 = 6.7. 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.

Example 3: (𝑥+5) 2 25 + (𝑦−4) 2 16 =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 (5,4). 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 - c2 16 = 25 - c2; c2=9; c = 3. 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.

How to enter into a calculator. (𝑥+5) 2 25 + (𝑦−4) 2 16 =1 16(𝑥+5) 2 25 + (𝑦−4) 2 =16 Multiply by 16 to isolate y-term. (𝑦−4) 2 =16− 16(𝑥+5) 2 25 Subtract x-term from both sides. 𝑦−4=± 16− 16(𝑥+5) 2 25 Take the square root of both sides. 𝑦=4± 16− 16(𝑥+5) 2 25 Add 4 to both sides. 𝑦=4+{1,−1}∗ 16− 16 25 (𝑥+5) 2 Enter in y-editor of calculator.

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.

Hyperbola Terminology 2𝑎 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑎𝑥𝑖𝑠 2𝑏 =𝑙𝑒𝑛𝑔𝑡ℎ 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑥𝑖𝑠 (0,c) f1 Conjugate axis vertex (0,a) Transverse axis (0,b) (c,0) x (c,0) Center f1 x Center Conjugate axis f2 (a,0) (a,0) foci (b,0) (b,0) vertex (0, b) (0, a) Transverse axis 𝑏 2 = 𝑐 2 − 𝑎 2 f2 foci (0,c) ℎ, 𝑘 =𝑐𝑒𝑛𝑡𝑒𝑟 y  𝑏 𝑎 slope of asymptotes 𝑓 1 , 𝑓 2 =𝑓𝑜𝑐𝑖 y  𝑎 𝑏 slope of asymptotes 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 𝑦 2 𝑎 2 − 𝑥 2 𝑏 2 =1 𝑦−𝑘 2 𝑎 2 − 𝑥−ℎ 2 𝑏 2 =1 𝑥−ℎ 2 𝑎 2 − 𝑦−𝑘 2 𝑏 2 =1 𝑓𝑜𝑐𝑖 𝑎𝑟𝑒 𝑎𝑙𝑤𝑎𝑦𝑠 𝑜𝑛 𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒 𝑎𝑥𝑖𝑠 (𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠) 𝑎, 𝑏 𝑎𝑟𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑠 𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 ℎ,𝑘 𝑎 2 𝑖𝑠 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚 𝑏𝑒𝑓𝑜𝑟𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑠𝑖𝑔𝑛

Example 1: 𝑥 2 49 − 𝑦 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 = 7.3. Since foci are on transverse axis (horizontal in this case), plot 7.3 units left and right of center.

𝑦 2 36 − 𝑥 2 64 =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 - a2 64 = c2-36; c2=100; c = 10. Since foci are on transverse axis (vertical in this case), plot 10 units above and below the center.

Example 3: (𝑥−4) 2 25 − (𝑦−6) 2 36 =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 - a2 36 = c2-25; c2=61; c = 7.8. Since foci are on transverse axis (horizontal in this case), plot 7.8 units left and right of center.

How to enter into a calculator. (𝑥−4) 2 25 − (𝑦−6) 2 36 =1 36(𝑥−4) 2 25 − (𝑦−6) 2 =36 Multiply by 36 to isolate y-term. 36(𝑥−4) 2 25 − 36 = (𝑦−6) 2 Subtract constant from both sides and add y-term to both sides  36(𝑥−4) 2 25 − 36 =𝑦−6 Take the square root of both sides. 6 36(𝑥−4) 2 25 − 36 =𝑦 Add 6 to both sides. 𝑦=6= 1,−1 ∗ 36 25 (𝑥−4) 2 − 36 Enter in y-editor of calculator.

Find the equation of a parabola from a graph. Locate the vertex (0, 3) and a point that the parabola passes through (1, 6). Substitute values from above for h, k, x, and y into: 𝑦=𝑎 (𝑥−ℎ) 2 +𝑘 Solve for a and then re-write formula. 6=𝑎 (1−0) 2 +3 𝑎=3 𝑦=3 (𝑥−0) 2 +3 or 𝑦=3 𝑥 2 +3

Find the equation of a hyperbola from a graph. Locate the center (4, 6) and a point that the parabola passes through (10.25, 10.5). You will most likely be given that point. Determine a2 from graph. a=5 Since horizontal, substitute values from above for a, h, k, x, and y into: 0.5625= 20.25 𝑏 2 𝑥−ℎ 2 𝑎 2 − 𝑦−𝑘 2 𝑏 2 =1 10.25−4 2 5 2 − 10.5−6 2 𝑏 2 =1 𝑏 2 =36 1.5625− 20.25 𝑏 2 =1 𝑥−4 2 5 2 − 𝑦−6 2 6 2 =1 1.5625−1= 20.25 𝑏 2