Mathematics
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OpenStudy (anonymous):
How would u find the equation for this hyperbola-
4x^2-y^2=0 ????
13 years ago
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OpenStudy (oaktree):
That's not a hyperbola. It's a normal function. Add y^2: y^2=4x^2. Square root, and y=2x.
13 years ago
OpenStudy (amistre64):
+- ...
13 years ago
OpenStudy (amistre64):
its a degenerate hyperbola
13 years ago
OpenStudy (anonymous):
but i'm trying to make a hyperbola out of it..
13 years ago
OpenStudy (anonymous):
who you calling a degenerate?
13 years ago
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OpenStudy (amistre64):
ellen lol
13 years ago
OpenStudy (anonymous):
so how would i do it...
13 years ago
OpenStudy (anonymous):
ho ho ho
13 years ago
OpenStudy (amistre64):
y = +- 2x
13 years ago
OpenStudy (amistre64):
a degenerate hyperbola is an X shape
13 years ago
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OpenStudy (anonymous):
\[4x^2=y^2\]
\[y=2x \text{ or } y = -2x\] a nice pair of lines, what amistre said
degeneration X
13 years ago
OpenStudy (anonymous):
ok. that makes sense buthow would i graph it...
13 years ago
OpenStudy (amistre64):
with 2 lines; y = 2x and y=-2x
13 years ago
OpenStudy (anonymous):
k and the center would be the origin?
13 years ago
OpenStudy (amistre64):
well, both y intercepts amount to 0, so yes
13 years ago
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OpenStudy (anonymous):
cool. and would there be a foci and/or vertices?
13 years ago
OpenStudy (amistre64):
the verts touch and are at the same points; which is why this thing is degenerate; youd have to play about for the focuses tho; i cant be sure of them unless they are at the center as well
13 years ago
OpenStudy (anonymous):
k thank u sooooooo much! ur awesome!
13 years ago
OpenStudy (amistre64):
im awesome AND i smell like irish spring lol
13 years ago
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OpenStudy (anonymous):
haha random but awesome :)
13 years ago