Question

Please help!
Find the eccentricity of this equation:
(x-5)^2=4(x-5)

Answer 1

\[(y-5)^{2}=4(x-5)\]

Answer 2

start by figuring out the type of conic section the equation represents

Answer 3

i know that\[e=\frac{ c }{ a }\]
is it a parabola? because only one variable is squared when distributing?

Answer 4

Yes ! it is a parabola, and what do you know about eccentricity of parabola ?

Answer 5

http://www.mathsisfun.com/geometry/eccentricity.html

Answer 6

y=1 if its a parabola

Answer 7

thats right !!

Answer 8

so i set \[\frac{ c }{ a }\] equal to 1?

Answer 9

eccentricity of parabola is 1. we're done, right ?

Answer 10

what about c/a?

Answer 11

I think you can use that formula for ellipses/hyperbolas

Answer 12

\[(y-5)^{2}=4(x-5)\]
\[y ^{2}+25=4x-20\] how do i put it into one of those formulas?

Answer 13

oh so like (y+5)^2-4(x+20)=1 kind of thing?

Answer 14

\[(y+5)^{2}-4(x+20)=1\]

Answer 15

that gives me the center...can i find a from the center?

Answer 16

dont i need vertices to find a and b?

Answer 17

eccentricity = \(\large \dfrac{\text{distance between a point to Focus}}{\text{distance between a point to Directrix}}\)