Expert: Sherman D. - 5/11/2004

Question
1. write standard form of the equation of the circle with radius 6 and center at (0,0)

2. write equation of an ellipse, verticies are
(-2,0) and (2,0) and the co-verticies are (0,-8) and (0,8)

3. graph the circle x^2+y^2=16

4. write standard form of the equation of the circle that passes through point (6,-6) )(actually it may have been (1,-6)) and the center at the origin

5. graph (x^2/49)+(y^2/9)=1

6.write equation of ellipse, vertex is (9,0) co-vertex is (0,5) and the center is at (0,0)

7.write equation of ellipse, vertex is (5,0) focus is (4,0) and the center is at (0,0)

8. sketch the graph of (x^2/4)+(y^2/16)=1

thanks a lot again!

Answer
Sorry, i had trouble doing 4 and 7.

Lets see what i can do for you.

Websites that i used, www.quickmath.com, www.calculator.com, and http://colbycc.edu/www/math/algebra/conics.htm

1.) write standard form of the equation of the circle with radius 6 and center at (0,0)

formula is
x^2 + y^2 = r^2
whereas "r" is the radius.

ANS:

x^2 + y^2 = 36

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2.) write equation of an ellipse, verticies are
(-2,0) and (2,0) and the co-verticies are (0,-8) and (0,8)

Formula
((x^2)/(a^2)) + ((y^2))/(b^2)) = 1
Whereas "a" and "b" are the vertices

ANS : ((x^2)/4) + ((y^2)/64) = 1

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3.) graph the circle x^2 + y^2 = 16

For a graph, go to www.calculator.com, click Graphing, and type in sqrt(-(x^2) + 16), click enter, then type in -sqrt(-(x^2) + 16), click enter. It may look like an ellipse, but if you look, the points are equal distant from the origin on all sides.

or go to quickmath and type in your problem like i have drawn it.

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4.) write standard form of the equation of the circle that passes through point (6,-6) )(actually it may have been (1,-6)) and the center at the origin

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5.) graph (x^2/49)+(y^2/9)=1

for a graph, go to www.quickmath.com, and type in
((x^2)/49) + ((y^2)/9) = 1,
or

go to www.calculator.com, click graphing, and type in

y^2/9 = (-x^2)/49 + 1
y^2 = (-9/49)x^2 + 9
y^2 = 9((-1/49)x^2 + 1)
y = 3sqrt((-1/49)x^2 + 1)

Type in 3sqrt((-1/49)x^2 + 1), click enter, then type in -3sqrt((-1/49)x^2 + 1), since the site doesn't understand sqrt as being 2 solutions.

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6.)write equation of ellipse, vertex is (9,0) co-vertex is (0,5) and the center is at (0,0)

((x^2)/81) + ((y^2)/25) = 1

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7.)write equation of ellipse, vertex is (5,0) focus is (4,0) and the center is at (0,0)

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8.) sketch the graph of ((x^2)/4) + ((y^2)/16) = 1
For a graph, go to www.quickmath.com, click Plot(Under equations), type in what i have typed, and you should get vertices points at (2,0),(-2,0),(0,4)and(0,-4)

or go to www.calculator.com, click Graphing, and type in the ANS : i give you, since the website doesn't do xy problem, it can only do problems in y = form.

((x^2)/4) + ((y^2)/16) = 1
((y^2)/16) = (x^2/4) + 1
y^2 = 16(-((x^2)/4) + 1)
y^2 = 4x^2 + 16
y = sqrt(-4(x^2) + 16)
y = sqrt(-4(-x^2) + 4))
y = 2sqrt(-(x^2) + 4)

ANS : to type in 2sqrt(-(x^2) + 4), since the calculator doesn't understand unless its in y = form.