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!
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