Question

solve

Answer 1

i have doubt that statement is incorrect, can any body guide me, plse friends

Answer 2

i don't get anything

Answer 3

0,4

Answer 4

kk.i got it but one thing there are 2 asympotes , so there will be two equations formed

Answer 5

first one you have solved

Answer 6

i mean to say that you have solved this part
Since the asymptote is y = 1/2 x and y =( b/a) x is the equation of an asymptote, that means that b = 1 and a = 2
also i the question, there is mention,
the asymptote is y =- 1/2 x
where this part has gone

Answer 7

b is a distance, and a is a distance .... those distances are not 1 and 2

the ratio of b to a is 1:2

Answer 8

@msingh asymptotes are \[\LARGE y = \pm \frac{ b }{a }x\] so the negative one isn't needed to find a and b.

Answer 9

k

Answer 10

thank you so much friends @agent0smith and @amistre64

Answer 11

your welcome, but you do understand that \(b\ne1,a\ne2\) right?

Answer 12

yes

Answer 13

then can you tell me what b and a are equal to?

Answer 14

b=1
and
a=2

Answer 15

no

Answer 16

but mertsj said and has already proved

Answer 17

b is a distance, a specific value; it measure the distance from the center to a vertex.

the center is at (0,0) and a vertex is at (0,2) what is the distance between those points?

Answer 18

the ratio of b to a is\[\frac{b}{a}=\frac12\]
sicne the distance from center to vertex is "2", then b=2

\[\frac{2}{a}=\frac12\]therefore a = 4

Answer 19

what mertsj has done, is that way not correct

Answer 20

http://www.wolframalpha.com/input/?i=%28y%2F1%29%5E2-%28x%2F2%29%5E2+%3D+1%2C%28y%2F2%29%5E2-%28x%2F4%29%5E2+%3D+1

Answer 21

hmmm, i could be off :) been a long day

Answer 22

a hyperbola is an inverted ellipse, so there is a characteristic ellipse that can be associated with it that has the same b a values

Answer 23

the y-x setup thru me :)

Answer 24

for starters, we know that when x=0, y=2, so (y/2)^2 becomes (2/2)^2 = 1

Answer 25

\[(\frac y2)^2-(\frac x4)^2=1\]
has asymptotes of \[y=\pm\frac{1}{2}x\]

http://www.wolframalpha.com/input/?i=%28y%2F2%29%5E2-%28x%2F4%29%5E2+%3D+1

Answer 26

\[(\frac y1)^2-(\frac x2)^2=1\]
has the same asymptotes, but its vertexs are at 1 and -1

Answer 27

also to note, in the asymptote equation; b/a is a slope defined usually as y/x

so b has to associate with y in this case.

Answer 28

..... odd, ive given ample "other" sources to back the validity of my idea.

the hyperbola equation\[\frac{y^2}{1^2}-\frac{x^2}{2^2}=1\]when x=0 has verts
\[\frac{y^2}{1^2}-\frac{0^2}{2^2}=1\]
\[\frac{y^2}{1^2}=1\](0,1) and (0,-1)
which are not the given verts in the problem ....., yet I have used the 1 and 2 from the asymptote

Answer 29

now \[\frac{y^2}{2^2}-\frac{x^2}{1^2}=1\]does have verts at (0,-2) and (0,2)
but its asymptotes are y = +-2x

Answer 30

http://www.wolframalpha.com/input/?i=%28y%2F2%29%5E2-%28x%2F1%29%5E2+%3D+1

Answer 31

given that \[slope:\frac ba~relates~to~\frac yx\]b is under y
a is under x

a = 1

Answer 32

b=2, a=1 that is

Answer 33

thats misleading tho ....

Answer 34

"for hyperbolas opening up/down, the asymptotes are: \(\Large y = \frac abx\)" so they corrected the matter

Answer 35

its given that\[asymp:y=\frac bax\]
divide both sides by x
\[\frac yx=\frac ba \]in this problem

Answer 36

..... i see what i might be overlooking

Answer 37

the asymptotes of\[\frac{y^2}{2^2}-\frac{x^2}{1^2}=1\\~\\y=\pm2x\]

http://www.wolframalpha.com/input/?i=%28y%2F2%29%5E2-%28x%2F1%29%5E2+%3D+1

just by convention of using y/x as slopes regardless of a and b

Answer 38

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

Answer 39

\[\frac{y^2}{2^2}-\frac{x^2}{4^2}=1\]
\[\frac{y^2}{2^2}=(\frac{x}{4})^2+1\]
\[y^2=(\frac{2}{4}x)^2+4\]
\[y^2=(\frac{1}{2}x)^2+4\]
this has a slope of 1/2

Answer 40

i dont mind being wrong :) i live in error most of the time .... but im sure of this issue here

Answer 41

lol, now i just feel like your saying that to shut me up ;)

Answer 42

so, vat's an answer -exact

Answer 43

centre is (0,0)
a= 4
b=2
that 's it

Answer 44

@amistre64 and @msingh a is usually taken as the distance from the center to the vertices. So a=2 here. Then b=4, from the asymptote equation (which i might've given incorrectly earlier, should be y = (a/b) x)

see here: http://www.mathwarehouse.com/hyperbola/images/picture-of-hyperbola-vertical-transverse.gif

Answer 45

what really bites about all this, is that a and b are just generic place holders for constructing the hyperbola. The equation for the hyperB was (y/2)^2 - (x/4)^2 = 1 regardless of what value we wanted to call a or b .....

but yes, since the "proper?" construct of an up and down has asymptotes as: a/b , which then relates to 1/2 .... then a/b = 2/4 is the "solution" :/

Answer 46

Yes, it doesn't matter, the only thing that matters is the value under the positive term (x or y) is the distance from the center to the vertex (squared).