SOLUTION: Find the four corners of the fundamental rectangle of the hyperbola: (x^2/81)-(x^2/36)=1

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Question 334131: Find the four corners of the fundamental rectangle of the hyperbola: (x^2/81)-(x^2/36)=1
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

There was a typo of x for y, in one of the two places, so
I had to do both cases, since I couldn't tell whether it 
was meant to be:

x%5E2%2Fa%5E2+-+y%5E2%2Fb%5E2%22%22=%22%221

or

y%5E2%2Fa%5E2+-+x%5E2%2Fb%5E2%22%22=%22%221

---------------------------------------------------

So both cases are below:

The two rules needed are:

1. The four corners of the fundamental rectangle of the hyperbola

x%5E2%2Fa%5E2+-+y%5E2%2Fb%5E2%22%22=%22%221 which opens right and left are

are (a,b), (a,-b), (-a,b),(-a,-b)

2. The four corners of the fundamental rectangle of the hyperbola

y%5E2%2Fa%5E2+-+x%5E2%2Fb%5E2%22%22=%22%221 which opens upward and downward

are (b,a), (b,-a), (-b,a),(-b,-a)

Therefore,

1. The four corners of the fundamental rectangle of the hyperbola

x%5E2%2F81+-+y%5E2%2F36%22%22=%22%221 which opens right and left are

are (9,6), (9,-6), (-9,6),(-9,-6)

2. The four corners of the fundamental rectangle of the hyperbola

y%5E2%2F81+-+x%5E2%2F36%22%22=%22%221 which opens upward and downward

are (6,9), (6,-9), (-6,9),(-6,-9)

Edwin