Asked by Emily
Find the vertex, focus, and axis of symmetry of the parabola x^2-10x-8y+29 = 0.
==> I do it over and over and over, but I keep getting the same answers of (5,4) for the vertex, (5,6) for the focus, and x = 5 for the LOS, but according to this one website, everything except for the LOS is wrong. This is how I got the equation into vertex form:
x^2-10x+25 = 8y-29+25
(x-5)^2=8y-4
y = 1/8(x-5)^2+4
I don't see what I'm doing wrong -- can someone help me please? Thanks!! :)
==> I do it over and over and over, but I keep getting the same answers of (5,4) for the vertex, (5,6) for the focus, and x = 5 for the LOS, but according to this one website, everything except for the LOS is wrong. This is how I got the equation into vertex form:
x^2-10x+25 = 8y-29+25
(x-5)^2=8y-4
y = 1/8(x-5)^2+4
I don't see what I'm doing wrong -- can someone help me please? Thanks!! :)
Answers
Answered by
Emily
Ohh never mind, I found my error. I feel kinda stupid now though lol
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