Tutors Answer Your Questions about Quadratic-relations-and-conic-sections (FREE)
Question 144994: The vertex of the parabola below is at the point (1, 3) and the point (2, 6) is on the parabola. What is the equation of the parabola?
i had gotten y = 6(x - 1)^2 + 3 but it came out wrong.
these are the other options
A. y = 3(x + 1)^2 - 3
C. y = 3(x - 1)^2 + 3
D. x = 2(y - 3)^2 + 1
Click here to see answer by jim_thompson5910(35256)  |
Question 144994: The vertex of the parabola below is at the point (1, 3) and the point (2, 6) is on the parabola. What is the equation of the parabola?
i had gotten y = 6(x - 1)^2 + 3 but it came out wrong.
these are the other options
A. y = 3(x + 1)^2 - 3
C. y = 3(x - 1)^2 + 3
D. x = 2(y - 3)^2 + 1
Click here to see answer by stanbon(75887) |
Question 144997: In which direction does the hyperbola given by the equation below open? Check all that apply.
(y+7)^2/3^2 - (x-4)^2/7^2=1
i had originally put down but it came out wrong.
these are the possible awnsers...
A. Down
B. Up
C. Right
D. Left
Click here to see answer by stanbon(75887) |
Question 144997: In which direction does the hyperbola given by the equation below open? Check all that apply.
(y+7)^2/3^2 - (x-4)^2/7^2=1
i had originally put down but it came out wrong.
these are the possible awnsers...
A. Down
B. Up
C. Right
D. Left
Click here to see answer by shahid(44) |
Question 145167: Find the vertex and focus of the parabola given by
x^2 = -1/8y
This is what I tried:
4p = -1/8
then divide by 4 so
p = - 1/32
vertex = (0,0,)
focus = (0, -1/32)
Please let me know if I did this right, thanks!
Click here to see answer by stanbon(75887) |
Question 145137: Find the equation of an ellipse with its center at (1,2), focus at (6, 2) and containing the point (4,6).
I have so far (x-1)^2 + (y-2)^2 = 1, but I cannot figure out what my a^2 and b^2 should be for my denominators. Please help.
Click here to see answer by Edwin McCravy(20059)  |
Question 145751: Help I am trying to fully understand how to find the real solutions of the system. I have no problem solving a problem like : x^2+2y^2=12 and 3x^2-y^2=8 but when the problem looks like this: xy+6=0 and x-y=-5 I cant figure out where to start. Your help would do a lot for me thanks.
Click here to see answer by oscargut(2103)  |
Question 147195: Write an equation in standard form for:
1. An ellipse with center a the origin, one vertex at (0,5) and one co-vertex at (0,2).
2. A parabola with vertex at the origin and directrix y=-2.
3. A circle with center (0,0) passing through (-3,4).
4. Hyperbola with vertices (8,-4) and (8,4) and foci (8,-6) and (8,6).
A few basic questions: What is a directrix and what is foci?
The other question: How would I go about figuring out the equations? What steps should I take?
Click here to see answer by Edwin McCravy(20059) |
Question 147808: My question is: Write the equation for the ellipse whose center is the origin, has a horizontal major axis length of 12 and passses through the point (-4,2).
I need help with this, since the horizontal axis is at the origin I get that the foci is (-6,0) and (6,0). I find that the equation will be  but I am having trouble finding what  equals. Can you help me?
Click here to see answer by scott8148(6628)  |
Question 147808: My question is: Write the equation for the ellipse whose center is the origin, has a horizontal major axis length of 12 and passses through the point (-4,2).
I need help with this, since the horizontal axis is at the origin I get that the foci is (-6,0) and (6,0). I find that the equation will be but I am having trouble finding what equals. Can you help me?
Click here to see answer by Edwin McCravy(20059) |
Question 148000: Find the vertex and intercepts for this parabola:
g(x) = x^2 + x - 6
This is how I have worked it:
a = 1, b = 1, c = -6
x = -b/2a
x = -1/[2(1)] = -1/2
g(-1/2) = (-1/2)^2 + (-1/2) - 6
= 25/2
So, the vertex is (-1/2, 25/2)
Am I correct so far?
How do I verify that I am correct? What should my next step be?
I appreciate your help very much! Thank you!
Click here to see answer by jim_thompson5910(35256) |
Question 148125: Could someone please help me with this question?
Thanks in advance..
Michael
The edge of my kitchen counter top is the shape of an ellipse. Since my tiles were 18" squares, the deepest part is 18 inches deep and the edge tapers to 12 inches deep. The counter will be 6 feet long. You can see the counter outlined below if you look at the area that is both inside the blue ellipse and inside the brown rectangle at the same time.
What is the equation of this ellipse if everything is figured in inches?
Where do the foci of my “tacks” need to be placed to draw my counter edge?
How long should the “string” be to generate this curve?
Click here to see answer by stanbon(75887) |
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