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 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 𝑦 2 𝑏 2 =1
𝑦 2 𝑎 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 𝑦−𝑘 2 𝑏 2 =1
𝑦−𝑘 2 𝑎 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

4. Example 1:
𝑥 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 =  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:
𝑥 𝑦 =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.

6. Example 3:
(𝑥+5) (𝑦−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 (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 - c = 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.

7. How to enter into a calculator.
(𝑥+5) (𝑦−4) =1
16(𝑥+5) (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− (𝑥+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
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

10. Example 1:
𝑥 − 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.

11. 𝑦 − 𝑥 =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 = 10. Since foci are on transverse axis (vertical in this case), plot 10 units above and below the center.

12. Example 3:
(𝑥−4) − (𝑦−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 = 7.8. 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.
(𝑥−4) − (𝑦−6) =1
36(𝑥−4) − (𝑦−6) 2 =36
Multiply by 36 to isolate y-term.
36(𝑥−4) − 36 = (𝑦−6) 2
Subtract constant from both sides and add y-term to both sides
 36(𝑥−4) − 36 =𝑦−6
Take the square root of both sides.
6 36(𝑥−4) − 36 =𝑦
Add 6 to both sides.
𝑦=6= 1,−1 ∗ (𝑥−4) 2 − 36
Enter in y-editor of calculator.

14. 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) or
𝑦=3 𝑥 2 +3

15. 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= 𝑏 2
𝑥−ℎ 2 𝑎 2 − 𝑦−𝑘 2 𝑏 2 =1
10.25− − − 𝑏 2 =1
𝑏 2 =36
1.5625− 𝑏 2 =1
𝑥− − 𝑦− =1
1.5625−1= 𝑏 2