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/
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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(13206) (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:
----------------------------------------------
(revised presentation, after question from student....)
When x = -2, the y value is 1:
Then a/b=4 leads to b=1/2.
That gives you the correct equation.
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
from the graph you see that you need equation:
center (,)=(,)
the length of the transverse axis is the distance between vertices and you see 0n the graph that =>
so far equation is
I picked point (, ) which lie on one asymptote
the equations of the asymptotes are = ±
= ±
= ±
=>
or
=>
and, your equation is:
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