SOLUTION: Determine the equation of the hyperbola in standard form and general form. https://i.ibb.co/4jjgzVK/1.jpg Answer for the standard equation was ((y + 1)^2)/4 - ((x + 2)^2)/(1/

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Determine the equation of the hyperbola in standard form and general form. https://i.ibb.co/4jjgzVK/1.jpg Answer for the standard equation was ((y + 1)^2)/4 - ((x + 2)^2)/(1/      Log On


   

Question 1204698: Determine the equation of the hyperbola in standard form and general form.
https://i.ibb.co/4jjgzVK/1.jpg
Answer for the standard equation was ((y + 1)^2)/4 - ((x + 2)^2)/(1/4) = 1
My answer was of course wrong: ((y + 1)^2)/16 - ((x + 2)^2)/1 = 1
There was no point in rewriting into general form when my standard form was already incorrect
What I did:
Center (-2, -1)
I picked points (-1, 3) and (-3, 3) on the two asymptotes to find slope +/- 4
Does that mean a = 4 and b = 1?
Because that’s what I used to plug into the equation ((y - k)^2)/a^2 - ((x - h)^2)/b^2 = 1
Found 2 solutions by greenestamps, MathLover1:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The slopes of the asymptotes being 4 and -4 does not mean a=4 and b=1. It only means a/b=4.

The given graph shows the larger y value when x=-2 has to be 1. Use that to find a and then b:

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(revised presentation, after question from student....)
%28y%2B1%29%5E2%2Fa%5E2-%28x%2B2%29%5E2%2Fb%5E2=1

When x = -2, the y value is 1:

%281%2B1%29%5E2%2Fa%5E2-%28-2%2B2%29%5E2%2Fb%5E2=1
%282%29%5E2%2Fa%5E2-%280%29%5E2%2Fb%5E2=1
4%2Fa%5E2=1
a%5E2=4
a=2

Then a/b=4 leads to b=1/2.

That gives you the correct equation.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
from the graph you see that you need equation:
%28y+-+k%29%5E2%2Fa%5E2+-+%28x+-+h%29%5E2%2Fb%5E2+=+1
center (h,k)=(-2,-1)
the length of the transverse axis is the distance between vertices and you see 0n the graph that 2a=4 => a=2
so far equation is
%28y+-k%29%5E2%2Fa%5E2+-+%28x-h%29%5E2%2Fb%5E2+=+1
%28y+%2B1%29%5E2%2F4+-+%28x%2B2%29%5E2%2Fb%5E2+=+1

I picked point (-1, -5) which lie on one asymptote
%28-5+%2B1%29%5E2%2F4+-+%28-1%2B2%29%5E2%2Fb%5E2+=+1
16%2F4+-+1%2Fb%5E2+=+1
16%2F4+-+1+=+1%2Fb%5E2
16%2F4+-+4%2F4+=+1%2Fb%5E2
b%5E2+=+1%2F3+
b+=+1%2Fsqrt%283%29+

the equations of the asymptotes are y= ± %28a%2Fb%29%28x-h%29%2Bk
y= ± %282%2F1%2Fsqrt%283%29%29%28x%2B2%29-1
y= ± 2sqrt%283%29%28x%2B2%29-1

y=+2sqrt%283%29%28x%2B2%29-1 => y=2sqrt%283%29+x+%2B+4sqrt%283%29+-+1
or
y=+-+2sqrt%283%29%28x%2B2%29-1=> y=-2sqrt%283%29+x+-+4sqrt%283%29+-+1

and, your equation is:

%28y+-+%28-1%29%29%5E2%2F2%5E2+-+%28x+-+%28-2%29%29%5E2%2F%281%2Fsqrt%283%29%29%5E2+=+1

%28y+%2B1%29%5E2%2F4+-+%28x%2B2%29%5E2%2F%281%2F3%29+=+1

%28y+%2B1%29%5E2%2F4+-+3%28x%2B2%29%5E2+=+1