Question

Can someone please help me find the foci, vertices, and asymptotes of \[\frac{ (x-4)^2 }{ 36 }-\frac{ (y-2)^2 }{ 9 }=1\]

Answer 1

@Hero @agent0smith @dan815 can any of you help me with this? :)

Answer 2

center you read off right from the equation \[\frac{ (x-4)^2 }{ 36 }-\frac{ (y-2)^2 }{ 9 }=1\]looks like \[\frac{ (x-h)^2 }{ a^2 }-\frac{ (y-k)^2 }{ b^2 }=1\]where the center is \((h,k)\) so your center is \((4,2)\)
also your \(a^2=36\) and so \(a=6\)
also \(b^2=9\) making \(b=3\)

Answer 3

I have that the center is (4,2)
Eccentricity is √3/2
Length of Transverse Axis is 12
Length of Conjugate Axis is 6 but I need help finding the foci, vertices, and asymptotes

Answer 4

I have no idea what the formulas are for those 3 or how to find them :(

Answer 5

before we can find the foci, we need to know what it looks like
do you know how it is oriented?

Answer 6

what do you mean?

Answer 7

like the first one

Answer 8

ok good
that tells you that the vertices are to the right and left of the center (not above and below)

Answer 9

oh okay, now what do i do? :)

Answer 10

and now it is easy, because you have \(a^2=36\) making \(a=6\)
go \(6\) units to the right and left of \((4,2)\)

Answer 11

sooo would that be (-6,2) and (6,2)??

Answer 12

no, 6 units to the right and left of \((4,2)\) not to the right and left of \((0,2)\)

Answer 13

in simpler english, add and subtract \(6\) to the first coordinate \(4\)

Answer 14

clear or no?

Answer 15

oh so (-10,2) and (10,2)?

Answer 16

not quite
\(4+6=10\) so the \((10,2)\) is right

Answer 17

how about the second one?

Answer 18

ahh, so would it be -2,2?

Answer 19

yes it would!

Answer 20

thanks! so that is the foci?

Answer 21

now we have the vertices, namely \((-2,2)\) and \((10,2)\)
next are the foci

Answer 22

oh!

Answer 23

no that was the vertices
we do the foci now

Answer 24

again we are going to go the same number of units to the left and right of the center, but this time we need \(c\) which is found via \(c^2=a^2+b^2\) just like in pythagoras
since \(a^2=36\) and \(b^2=9\) we have \(c^2=36+9=45\) and so \(c=\sqrt{45}\)

Answer 25

your teacher probably wants you to write \(\sqrt{45}\) in simplest radical form as \(3\sqrt{5}\)

Answer 26

yes! like that, so that would be the vertices or is there more to it?

Answer 27

you have to add and subtract that from the first coordinate of the center, just like we did for the vetices
but in this case you just and and subtract by putting a plus and minus sign there

Answer 28

in other words they look like
\[(4-3\sqrt5,2),(4+3\sqrt5,2)\]

Answer 29

would that be final or do i actually subtract and add that?

Answer 30

just leave it like that
you cannot actually add or subtract unless you take a calculator and find a decimal approximation for those number
i would leave them be

Answer 31

i guess all that is left is the asymptotes right?

Answer 32

thank you so much!
& yes please :)

Answer 33

for that you need the slope(s)

Answer 34

and that is also pretty easy, since the slopes are \(\pm\frac{b}{a}\) which in your case is \(\pm\frac{6}{3}=\pm 2\)

Answer 35

whoa whoa typo there sorry

Answer 36

it is \(\pm\frac{b}{a}\) all right, but your \(a=6\) and \(b=3\) i had it backwards
so your slopes are \(\pm\frac{3}{6}=\pm\frac{1}{2}\)

Answer 37

its okay! lol

Answer 38

then the lines must go through the center, \((4,2)\) so by the point slope formula, if the slope is \(\frac{1}{2}\) then the line is
\[y-2=\frac{1}{2}(x-4)\] or solving for \(y\) you get
\[y=\frac{1}{2}x\]

Answer 39

if the slope is \(-\frac{1}{2}\) you get
\[y-2=-\frac{1}{2}(x-4)\] solve for \(y\) and get
\[y=-\frac{1}{2}x+4\]

Answer 40

if you need another worked out example with a good explanation, google "hyperbola purplemath"

Answer 41

http://www.purplemath.com/modules/hyperbola2.htm

Answer 42

so it would just by a vertical asymptote of y=-1/2x+4??

Answer 43

they do not have vertical asymptotes, they have two slant asymptotes
you have to write two of them
one is \(y=\frac{1}{2}x\) the other is \(y=-\frac{1}{2}x+4\) both are asymptotes, neither are vertical