Question

find the vertices and foci of the equation:

Answer 1

\[\frac{ (x-1)^2 }{ 4}-\frac{ (y+2)^@ }{ 16 }=1\]

Answer 2

whats your attempt ?

Answer 3

hmm i figure that its a hyperbola isn't? and the vertices is +_4?

Answer 4

yes this is a hyperbola,
suppose the question would have been
x^2/4 - y^2 /16= 1

then can you find the co-ordinates of focii ?

Answer 5

hmm i'll try my guess would be \[+-2\sqrt{5}\] i'm wrong isn't it >,<

Answer 6

focii is (ae,0) and (-ae,0) right ?

Answer 7

i think is should be the y because the denominator of y it larger than the x

Answer 8

I am sorry but am getting confused here! :O
really sorry..

Answer 9

anyways, what is the eccentricity of the curve ?

Answer 10

oh its ok i'm just looking for some opinion, hmm the eccentricity would be \[\frac{ \sqrt{5} }{ 2 }\] i guess

Answer 11

and I don;t think denom of y would make any diff since co=efficient of y^2 is negative and x^2 is positive .

Answer 12

and eccentricity should be sqrt5 only. I guess.
its sqrt(1 + b^2/a^2) right ?

Answer 13

hmm i wonder about that i don't think that even if we expand the binomial (y+2)^2 the y^2 term would be positive and the x^2 term as well

Answer 14

hmm i wonder about that isn't it that the formula for eccentricity is\[e=\frac{ c }{ a }\]

Answer 15

yes, that is
but c = sqrt(b^2 + a^2) for hyperbola .

Answer 16

then that would result to sqrt5 isn't it

Answer 17

yes, sqrt5 is your eccentricity.
which would mean the foci are (2sqrt5,0) and (-2sqrt5,0)

This is for simple hyperbola, now we will have to make transformations.

You'll see the x co-ordinate needs to be shifted by 1 unit to the right, and y co-ordinate 2 units down.
which would mean new foci are
(1+ 2sqrt5 ,-2) and (1-2sqrt5 ,-2)

Please tell me if you follow ?

Answer 18

i think sqrt5 would have to be divide to 2, hmm why we need to shift?

Answer 19

This gives some insight
https://www.purplemath.com/modules/hyperbola.htm

the "center" is given by the # 's in x-1 and y+2 : the center is 1,-2

we are subtracting y^2 so this is a parentheses shape
(subtracting x^2 gives a smile/frown)

the vertices are ± a away (in the x direction) where a^2 divides the x^2 term

the foci are at ±c from the center (in the x direction)
where c^2= a^2 + b^2

Answer 20

in a hyperbola, the a^2 always divides the positive term. in this case the x^2 term