Find the equation of the hypeerbola whose axes are along the coordinate axes and which passes through `(-3,4), and (5,6)`

Answer

To find the equation of the hyperbola whose axes are along the coordinate axes and which passes through the points \((-3, 4)\) and \((5, 6)\), we can follow these steps:

### Step 1: Write the standard form of the hyperbola
The standard equation of a hyperbola with its axes along the coordinate axes is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the first point \((-3, 4)\)
Substituting the point \((-3, 4)\) into the hyperbola equation:
\[ \frac{(-3)^2}{a^2} - \frac{4^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} - \frac{16}{b^2} = 1 \]
Multiplying through by \(a^2b^2\) to eliminate the denominators gives:
\[ 9b^2 - 16a^2 = a^2b^2 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point \((5, 6)\)
Now, substitute the point \((5, 6)\) into the hyperbola equation:
\[ \frac{5^2}{a^2} - \frac{6^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{25}{a^2} - \frac{36}{b^2} = 1 \]
Again, multiplying through by \(a^2b^2\) gives:
\[ 25b^2 - 36a^2 = a^2b^2 \quad \text{(Equation 2)} \]

### Step 4: Set the two equations equal to each other
Now we have two equations:
1. \(9b^2 - 16a^2 = a^2b^2\)
2. \(25b^2 - 36a^2 = a^2b^2\)

We can set the left-hand sides of these equations equal to each other:
\[ 9b^2 - 16a^2 = 25b^2 - 36a^2 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 36a^2 - 16a^2 = 25b^2 - 9b^2 \]
This simplifies to:
\[ 20a^2 = 16b^2 \]
Dividing both sides by 4 gives:
\[ 5a^2 = 4b^2 \]
Thus, we can express \(a^2\) in terms of \(b^2\):
\[ a^2 = \frac{4}{5}b^2 \]

### Step 6: Substitute \(a^2\) back into Equation 1
Substituting \(a^2 = \frac{4}{5}b^2\) into Equation 1:
\[ 9b^2 - 16\left(\frac{4}{5}b^2\right) = \left(\frac{4}{5}b^2\right)b^2 \]
This simplifies to:
\[ 9b^2 - \frac{64}{5}b^2 = \frac{4}{5}b^4 \]
Finding a common denominator (which is 5):
\[ \frac{45b^2 - 64b^2}{5} = \frac{4}{5}b^4 \]
This simplifies to:
\[ \frac{-19b^2}{5} = \frac{4}{5}b^4 \]
Multiplying through by 5:
\[ -19b^2 = 4b^4 \]
Rearranging gives:
\[ 4b^4 + 19b^2 = 0 \]
Factoring out \(b^2\):
\[ b^2(4b^2 + 19) = 0 \]

### Step 7: Solve for \(b^2\)
Since \(b^2\) cannot be negative, we have:
\[ 4b^2 + 19 = 0 \quad \text{(no real solutions)} \]
This indicates that there are no real solutions for \(b^2\), which implies that the hyperbola does not exist for the given points.

### Final Equation
Thus, the equation of the hyperbola is:
\[ \frac{x^2}{-\frac{19}{5}} - \frac{y^2}{-\frac{19}{4}} = 1 \]
This leads to:
\[ 4y^2 - 5x^2 = 19 \]