What is the circumference of a circle with an area of 16π in²
1 answer:
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Answer:
- (x-4.5)^2 +(y +5)^2 = 30.25
- x = (1/8)y^2 +(1/2)y +(1/2)
- y^2/36 -x^2/64 = 1
- x^2/16 +y^2/25 = 1
Step-by-step explanation:
1. Complete the square for both x and y by adding a constant equal to the square of half the linear term coefficient. Subtract 15, and rearrange to standard form.
(x^2 -9x +4.5^2) +(y^2 +10y +5^2) = 4.5^2 +5^2 -15
(x -4.5)^2 +(y +5)^2 = 30.25 . . . . . write in standard form
Important features: center = (4.5, -5); radius = 5.5.
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2. To put this in the form x=f(y), we need to add 8x, then divide by 8.
x = (1/8)y^2 +(1/2)y +(1/2)
Important features: vertex = (0, -2); focus = (2, -2); horizontal compression factor = 1/8.
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3. We want y^2/a^2 -x^2/b^2 = 1 with a=36 and b=(36/(3/4)^2) = 64:
y^2/36 -x^2/64 = 1
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4. In the form below, "a" is the semi-axis in the x-direction. Here, that is 8/2 = 4. "b" is the semi-axis in the y-direction, which is 5 in this case. We want x^2/a^2 +y^2/b^2 = 1 with a=4 and b=5.
x^2/16 +b^2/25 = 1
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The first attachment shows the circle and parabola; the second shows the hyperbola and ellipse.
