Question

Determine the asymptotes for the second degree equation 7x^2 +42x +6y+81=3y^2

Answer 1

I'm not at all sure that the equation you've typed in has any asymptotes. It's not a rational function. Except in trigonometry, only rational functions have asymptotes. Check to ensure that you've copi

Answer 2

copied this problem down correctly.

Answer 3

It is a trigonometry problem. I did copy down correctly. I know it is an equation to a hyperbola and has to be manipulated. I am pretty sure it turns into to 7x^2+42x =3y^2-6y. but I am not sure where the 81 goes

Answer 4

Oh, I see. In that case you mean "slant asymptotes," not vertical or horizontal ones.

Answer 5

Taking the equation you've posted, re-write it as follows:\[7x^2+42x =3y^2-6y \rightarrow 7(x^2+6x)=3(y^2-2y)\]

Answer 6

... as the first step towards putting this equation into "standard form" for the eq'n of a hyperbola.

Answer 7

You will have to "complete the square" twice, starting with each of the expressions in each set of parentheses. Are you able to do this?

Answer 8

Yes. When you complete the square you would get 7(x^2+6x+9) +3(y^2-2y+4)=81. It would then be 7(x+3)^2 + 3(y-2)^2=8. Where do you go from there?

Answer 9

\[7(x^2+6x)=3(y^2-2y)\]
Now move all terms to the left and side:
\[7(x^2+6x)-3(y^2-2y)=0\]
Complete the square for each parenthesis,
\[7(x^2+6x+9)-63-3(y^2-2y+1)+3=0\]
\[7(x^2+6x+9)-3(y^2-2y+1)=60\]
\[7(x+3)^2-3(y-1)^2=60\]
Now divide by 60 on both sides,
\[\frac{(x+3)^2}{60/7}-\frac{(y-1)^2}{20}=1\]
Now the equation is in the standard form

Answer 10

Last, you need to determine whether this hyperbola is a vertical or a horizontal one. Once you've done that, you can identify a^2 and b^2.

With acknowledgements to http://www.purplemath.com/modules/hyperbola.htm

See the illustration attached:

Answer 11

If you can identify your a and b correctly, the appropriate formula from the attached figure will help you identify the equations of your (slant) asymptotes.

Answer 12

Well this is a horizontal asymptote since the negative sign is on y. Now you just need to find the equation of the conjugate axis, which will give you the vertical asymptote

Answer 13

Sorry it is a horizontal hyperbola not a horizontal asymptote. A horizontal hyperbola has a vertical asymptote as far as I know

Answer 14

Comment #1: Yes, this is a horiz. hyperbola: It opens to the left and to the right.
Comment #2: The asymtopes of this hyperbola are SLANT, as I've mentioned before, not horiz. or vertical. Please see the illustration I posted a few minutes back.

Answer 15

Okay. This has been so much help. I think I got it. In Navk's problem the 81 was on the other side, so I do believe when I put the 81 in the amount on the other side should be 138. The equation then is \[(x+3)^{2}/138/7 + (y-1)^{2}/46=1\]

Answer 16

The asymtopes should then be \[y-1= \pm \frac{ \sqrt{46} }{ \sqrt{81/7} }(x+3)\]

Answer 17

You could check the accuracy of your asymptotes by graphing them. Do that if you have lots of free time, Otherwise, I'd suggest you move on to your next homework problem.

Answer 18

Okay. Perfect. Thank you!

Answer 19

:)