CH. 10 DAY 1

CH. 10: WE WILL STUDY: CONIC SECTIONS

WHEN A PLANE INTERSECTS A RIGHT CIRCULAR CONE, THE RESULT IS A CONIC

SECTION. THE FOUR TYPES ARE:

CIRCLEStandard Equation

if center is the origin

Standard Equationif (h,k) is the center

x2 +y2 = r2 (x-h)2 +(y-k)2 = r2

r = radius

CIRCLE

(x-h)2 +(y-k)2 = r2

Find the equation of a circle that has a center =(2,-1) and a radius equal to 3

(x-2)2 +(y- -1)2 = 32

(x-2)2 +(y+1)2 = 9

(x-h)2 +(y-k)2 = r2

CIRCLE

(x)2 +(y-3)2 = 52

Find the equation of a circle that has a center = (0,3) and a point on the circumference of the circle is (-4,6)

(x)2 +(y-3)2 = 25

(3)2 +(4)2 = r2

r = 5

First find the radius!

(x-h)2 +(y-k)2 = r2

CIRCLEFind the equation of a circle that has (2,-3) and (-10,-8) as the endpoints of a diameter.

(x+4)2 +(y+5.5)2=(6.5)2

Center: (-4,-5.5)

d =13 so r = 6.5

First find the center!

(5)2 +(12)2 = d2

Now find the radius

(x-h)2 +(y-k)2 = r2

(x+4)2 +(y+5.5)2=42.25

CIRCLEGiven the equation: 4(x-2)2+4(y+3)2 + 20 = 120A. Write the equation in Standard formB. Find the center of the circleC. Find the radius of the circle

(x-h)2 +(y-k)2 = r2Marker boards

4(x-2)2+4(y+3)2 + 20 = 120

4(x-2)2+4(y+3)2 = 100

A. (x-2)2+(y+3)2 = 25

C. Radius 5

B. Center(2,-3)

D. CHALLENGE: If the coefficients on the binomial terms were 4 and 5…how would this impact the equation of the circle?

D. It would not be a circle. It would be an ellipse!

CHANGE THE EQUATION FROM NON-STANDARD FORM TO STANDARD FORM

01046 22 yyxx

022 FEyDxCyBxyAx

(x-h)2 +(y-k)2 = r2

222222 )2()3(10)2(4)3(6 yyxx

4910)2()3( 22 yx

3)2()3( 22 yx

WRITE THE EQUATION IN STANDARD FORM

016844 22 yyx

(x-h)2 +(y-k)2 = r2

16844 22 yyx

5)1( 22 yx

16)2(4)0(4 22 yyx

To complete the square, send the “c” over the equal. Get the leading coefficient to be 1 by factoring out the 4. Then find the magic number using square of b/2

22222 )1(4)0(416))1(2(4)0(4 yyx

4016)1(4)0(4 22 yx

20)1(4)0(4 22 yx

Can I divide by “4” first thing?

In circles – yes – but not in some other conic sections so I wanted to get you ready for those!

CIRCLE

(x-h)2 +(y-k)2 = r2

Find the equation of a circle, in standard form, that has a center =(-2,-6) and is tangent to the line x=3

(x+2)2 +(y+6)2 = 52

(x+2)2 +(y+6)2 = 25

(x-h)2 +(y-k)2 = r2

CIRCLE

(x-h)2 +(y-k)2 = r2

Find the equation of a circle, in standard form, that has a center in quadrant 1 and is tangent to the lines x=-3, x=5, and the x-axis

(x-1)2 +(y-4)2 = 42

(x-1)2 +(y-4)2 = 16

(x-h)2 +(y-k)2 = r2

Center: (1,4)

Domain:

Range:

Circumference:

Area:

}53/{ xx}80/{ yx

8 dC 162 rA

HW: WS 10.1 DUE NEXT CLASS!

HW hints:

•#3 Divide by 3 first•#4 Add 8 first•Remember the right side of the equation is r2…not r•Simplify radicals if you can but do not convert them to decimals!