6-3 Conic Sections: EllipsesGeometric 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.

x

y

From each point in the plane, the sum of the distances to the foci is a constant.

Example:f1 f2

d2d1

fociPoint A: d1+d2 = c Point B: d1+d2 = c

BA

d1 d2

So, about those foci . . .

Center

y

f1f2foci

xMajor axis

Minor axis

Center

y

f1

f2

foci

x

Major axis

Minor axis

Ellipse Terminology

(c,0) (c,0)

(a,0)

vertex

(a,0)

(0,b)

(0, b)

(b,0)

vertex

(0, a)

(b,0)

(0,c)

(0,c)

(0,a)

+

+

+

+

𝑏2=𝑎2−𝑐2(h ,𝑘 )=𝑐𝑒𝑛𝑡𝑒𝑟𝑓 1 , 𝑓 2= 𝑓𝑜𝑐𝑖

𝑎 ,𝑏𝑎𝑟𝑒𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑠𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 (h ,𝑘 )𝑙𝑎𝑟𝑔𝑒𝑟 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑖𝑠𝑎2

𝑓𝑜𝑐𝑖𝑎𝑟𝑒𝑎𝑙𝑤𝑎𝑦𝑠 𝑜𝑛𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠(𝑙𝑜𝑛𝑔𝑒𝑟 𝑜𝑛𝑒)

|2𝑎|= h𝑙𝑒𝑛𝑔𝑡 𝑚𝑎𝑗𝑜𝑟 𝑎𝑥𝑖𝑠|2𝑏|= h𝑙𝑒𝑛𝑔𝑡 𝑚𝑖𝑛𝑜𝑟𝑎𝑥𝑖𝑠

+ Example 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: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: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.

Multiply by 16 to isolate y-term.+

Subtract x-term from both sides.

(𝑦−4 )2=16− 16 (𝑥+5)2

25

Take the square root of both sides.

𝑦=4±√16− 16(𝑥+5)2

25Add 4 to both sides.

𝑦−4=± √16− 16 (𝑥+5)2

25

Enter in y-editor of calculator.𝑦=4+{1 ,−1 }∗√(16−( 1625 )(𝑥+5)2)

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.

Center

y

f1f2 foci

x

Transverse axis

Conjugate axis

Center

y

f1

f2 foci

x

Transverse axis

Conjugate axis

Hyperbola Terminology

(c,0) (c,0)

(a,0)

vertex

(a,0)

(0,b)

(0, b)

(b,0)

vertex

(0, a)

(b,0)

(0,c)

(0,c)

(0,a)

𝑏2=𝑐2−𝑎2(h ,𝑘 )=𝑐𝑒𝑛𝑡𝑒𝑟𝑓 1 , 𝑓 2= 𝑓𝑜𝑐𝑖

𝑎 ,𝑏𝑎𝑟𝑒𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒𝑠𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 (h ,𝑘 )𝑎2𝑖𝑠 𝑖𝑛 𝑓𝑖𝑟𝑠𝑡𝑡𝑒𝑟𝑚𝑏𝑒𝑓𝑜𝑟𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠𝑖𝑔𝑛

𝑓𝑜𝑐𝑖𝑎𝑟𝑒𝑎𝑙𝑤𝑎𝑦𝑠 𝑜𝑛𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒𝑎𝑥𝑖𝑠(𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑒𝑠)

|2𝑎|= h𝑙𝑒𝑛𝑔𝑡 𝑡𝑟𝑎𝑛𝑠𝑣𝑒𝑟𝑠𝑒𝑎𝑥𝑖𝑠|2𝑏|= h𝑙𝑒𝑛𝑔𝑡 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒𝑎𝑥𝑖𝑠

slope of asymptotes slope of asymptotes

Example 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.

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: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.

Multiply by 36 to isolate y-term.

Subtract constant from both sides and add y-term to both sides

Take the square root of both sides.

Add 6 to both sides.

Enter in y-editor of calculator.

√ 36(𝑥−4 )225−   36=𝑦−6

6√ 36(𝑥−4 )225−   36=𝑦

𝑦=6= {1,−1 }∗√(( 3625 )(𝑥−4)2−   36)

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+𝑘Solve for a and then re-write formula.

6=𝑎(1−0)2+3𝑎=3

or 𝑦=3 𝑥2+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:

1 .5625− 20.25𝑏2

=1

1 .5625−1=20.25𝑏2

0 .5625=20.25𝑏2

𝑏2=36