Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

Help me please:

OpenStudy (anonymous):

Let e1 and e2 denote the eccentricites of the hyperbolas \[\frac{ x^2 }{ m^2 }-\frac{ y^2 }{ n^2 }=1 \] and \[\frac{ y^2 }{ n^2 }-\frac{ x^2 }{ m^2 }=1\] respectively. Verify that \[e \left(\begin{matrix}2 \\ 1\end{matrix}\right)e \left(\begin{matrix}2 \\ 2\end{matrix}\right)=e \left(\begin{matrix}2 \\ 1\end{matrix}\right)+e \left(\begin{matrix}2 \\ 2\end{matrix}\right)\]

OpenStudy (anonymous):

@triciaal

OpenStudy (anonymous):

@wio

OpenStudy (triciaal):

not current on this, sorry

OpenStudy (anonymous):

can you tag someone who might know?

OpenStudy (triciaal):

@satellite73, @mathstudent55, @jim_thompson5910 can you please help?

OpenStudy (anonymous):

by the way its "e sub 1 squared" incase you were confused

OpenStudy (triciaal):

@pico33 ?

OpenStudy (anonymous):

and eccentricity is basically c/a

OpenStudy (triciaal):

@jhannybean do you mind helping?

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

@Abhisar

OpenStudy (anonymous):

@dan815

OpenStudy (anonymous):

@Compassionate

OpenStudy (triciaal):

@ganeshie8 will you please help

OpenStudy (anonymous):

its actually \[e_1^2e_2^2=e_1^2+e_2^2\]

OpenStudy (anonymous):

not the other one i inserted

OpenStudy (jhannybean):

Not sure about these..

OpenStudy (anonymous):

do you know anyone who would know? and if you do can you tag them please?

OpenStudy (dan815):

draw what eccentricities are

OpenStudy (anonymous):

isnt it c/a?

OpenStudy (dan815):

whats c and a

OpenStudy (anonymous):

im trying to decide whether a is m or n

OpenStudy (anonymous):

c is \[\sqrt{m^2+n^2}\]

OpenStudy (dan815):

ok

OpenStudy (dan815):

you should google what eccentricities of hyprbole are

OpenStudy (anonymous):

what do you mean google it? isnt it just c/a?

OpenStudy (anonymous):

so like \[\frac{ m }{ \sqrt{m^2+n^2} }\]

OpenStudy (anonymous):

@wio

OpenStudy (dan815):

ok

OpenStudy (anonymous):

but is that e1 or e2?

OpenStudy (dan815):

lets say its e1, what is e2?

OpenStudy (anonymous):

n^2 over that number?

OpenStudy (anonymous):

so there is no way of knowing if "m" is "a" or "b"?

OpenStudy (dan815):

there is

OpenStudy (anonymous):

how?

OpenStudy (dan815):

if u learn what eccentricities are :P

OpenStudy (dan815):

im jk

OpenStudy (anonymous):

can you help me?

OpenStudy (anonymous):

its ok

OpenStudy (dan815):

but i think c/a c=the hyptenues of the major and miinor length

OpenStudy (dan815):

and a= probably the major

OpenStudy (anonymous):

isnt c the distance from the center to the vertex?

OpenStudy (dan815):

yeah that is equivalent to the hyp of m and n

OpenStudy (dan815):

eccentricity tells u how curved or close to parallel the hyperbola is

OpenStudy (anonymous):

so then how would you know which is "a" and "b"?

OpenStudy (dan815):

okay so think which one wud matter

OpenStudy (dan815):

say u have x^2/a^2-y^2/b^2=1 parabola

OpenStudy (dan815):

can u graph that

ganeshie8 (ganeshie8):

\[e_1 = \frac{c}{a} = \dfrac{\sqrt{m^2+n^2}}{m}\] \[e_2 = \frac{c}{a} = \dfrac{\sqrt{m^2+n^2}}{n}\]

OpenStudy (anonymous):

i dont remember how :(

ganeshie8 (ganeshie8):

plug them in the given equation and do ur algebra

OpenStudy (dan815):

|dw:1418718193757:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!