So I've drawn and graphed a picture on paper using the required conic sections for my math class, and I already know all of the formulas for parabola, hyperbola, circle, etc.
What I'm having trouble with is solving for y so I can put them into my graphing calculator and have them show up. Here is my horizontal ellipse equation in standard form:
x^2 / .25 + (y-9.25)^2 / .0625 = 1
For some reason, I can't solve for y; I keep getting stuck and it ends up not making sense. If anyone knows how to solve for y, please let me know! It would be much appreciated!
When I tried solving for y,
What I'm having trouble with is solving for y so I can put them into my graphing calculator and have them show up. Here is my horizontal ellipse equation in standard form:
x^2 / .25 + (y-9.25)^2 / .0625 = 1
For some reason, I can't solve for y; I keep getting stuck and it ends up not making sense. If anyone knows how to solve for y, please let me know! It would be much appreciated!
When I tried solving for y,
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I'll solve a similar conic for you; then follow this to get it in your calculator.
There's a BIG problem with ellipses, circles, and hyperbolas (other than the hyperbola xy = k). Their graphs are NOT functions! You should see this because a circle and an ellipse will ALWAYS fail the vertical line test...and hyperbolas that aren't asymptotic to the x and y axis will, too. (like x^2 - y^2 = 1.)
But you can divide it into TWO functions. Take the ellipse 4x^2 + y^2 = 16. You can solve for y as follows:
4x^2 + y^2 = 16
y^2 = 16 - 4x^2
y = +/- SQRT(16 - 4x^2)
Notice that this last expression (and actually NONE of the formulas above) give a function. If you put a value in for x, you probably won't get one unique value of y. Again, that's a vertical line failure. But we can graph it in a graphing calculator...you just need to graph TWO separate functions:
y1 = SQRT(16 - 4x^2)
y2 = -SQRT(16 - 4x^2)
The first function lies entirely above the x-axis. You'll get the top half.
The second function lies entirely below the x-axis. You'll get the (symmetric) bottom half.
If you have a (y - 1)^2 in there, it's not much different:
4x^2 + (y-1)^2 = 16
(y-1)^2 = 16 - 4x^2
y - 1 = +/- SQRT(16 - 4x^2)
y = 1 +/- SQRT(16 - 4x^2)
Again, two graphs. These graphs aren't symmetric about the line y=0 like before...but they ARE symmetric about the line y=1. You'll get:
y1 = 1 + SQRT(16 - 4x^2)
y2 = 1 - SQRT(16 - 4x^2)
The first graph is ALWAYS above (or touching) y=1, since SQRT of anything is nonnegative.
The second graph is ALWAYS below (or touching) y=1, since -SQRT of annything is nonpositive.
There's a BIG problem with ellipses, circles, and hyperbolas (other than the hyperbola xy = k). Their graphs are NOT functions! You should see this because a circle and an ellipse will ALWAYS fail the vertical line test...and hyperbolas that aren't asymptotic to the x and y axis will, too. (like x^2 - y^2 = 1.)
But you can divide it into TWO functions. Take the ellipse 4x^2 + y^2 = 16. You can solve for y as follows:
4x^2 + y^2 = 16
y^2 = 16 - 4x^2
y = +/- SQRT(16 - 4x^2)
Notice that this last expression (and actually NONE of the formulas above) give a function. If you put a value in for x, you probably won't get one unique value of y. Again, that's a vertical line failure. But we can graph it in a graphing calculator...you just need to graph TWO separate functions:
y1 = SQRT(16 - 4x^2)
y2 = -SQRT(16 - 4x^2)
The first function lies entirely above the x-axis. You'll get the top half.
The second function lies entirely below the x-axis. You'll get the (symmetric) bottom half.
If you have a (y - 1)^2 in there, it's not much different:
4x^2 + (y-1)^2 = 16
(y-1)^2 = 16 - 4x^2
y - 1 = +/- SQRT(16 - 4x^2)
y = 1 +/- SQRT(16 - 4x^2)
Again, two graphs. These graphs aren't symmetric about the line y=0 like before...but they ARE symmetric about the line y=1. You'll get:
y1 = 1 + SQRT(16 - 4x^2)
y2 = 1 - SQRT(16 - 4x^2)
The first graph is ALWAYS above (or touching) y=1, since SQRT of anything is nonnegative.
The second graph is ALWAYS below (or touching) y=1, since -SQRT of annything is nonpositive.