Find the of the hyperbola passing through the points (2, 1) and (4,3)

Answer

To find the equation of the hyperbola passing through the points (2, 1) and (4, 3), we will follow these steps:

### Step 1: Write the standard form of the hyperbola
The standard form of a hyperbola centered at the origin is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the first point (2, 1) into the equation
Substituting \(x = 2\) and \(y = 1\) into the hyperbola equation:
\[ \frac{2^2}{a^2} - \frac{1^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{4}{a^2} - \frac{1}{b^2} = 1 \quad \text{(Equation 1)} \]

### Step 3: Substitute the second point (4, 3) into the equation
Now, substituting \(x = 4\) and \(y = 3\) into the hyperbola equation:
\[ \frac{4^2}{a^2} - \frac{3^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{16}{a^2} - \frac{9}{b^2} = 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
We now have two equations:
1. \(\frac{4}{a^2} - \frac{1}{b^2} = 1\)
2. \(\frac{16}{a^2} - \frac{9}{b^2} = 1\)

To eliminate \(b^2\), we can manipulate these equations. From Equation 1, we can express \(\frac{1}{b^2}\):
\[ \frac{1}{b^2} = \frac{4}{a^2} - 1 \]

Substituting this into Equation 2:
\[ \frac{16}{a^2} - 9\left(\frac{4}{a^2} - 1\right) = 1 \]
Expanding this gives:
\[ \frac{16}{a^2} - \frac{36}{a^2} + 9 = 1 \]
Combining the terms:
\[ -\frac{20}{a^2} + 9 = 1 \]
Rearranging gives:
\[ -\frac{20}{a^2} = 1 - 9 \]
\[ -\frac{20}{a^2} = -8 \implies \frac{20}{a^2} = 8 \]
Thus, solving for \(a^2\):
\[ a^2 = \frac{20}{8} = \frac{5}{2} \]

### Step 5: Find \(b^2\)
Now substitute \(a^2 = \frac{5}{2}\) back into Equation 1 to find \(b^2\):
\[ \frac{4}{\frac{5}{2}} - \frac{1}{b^2} = 1 \]
This simplifies to:
\[ \frac{8}{5} - \frac{1}{b^2} = 1 \]
Rearranging gives:
\[ -\frac{1}{b^2} = 1 - \frac{8}{5} = -\frac{3}{5} \]
Thus:
\[ \frac{1}{b^2} = \frac{3}{5} \implies b^2 = \frac{5}{3} \]

### Step 6: Write the equation of the hyperbola
Now we have \(a^2 = \frac{5}{2}\) and \(b^2 = \frac{5}{3}\). The equation of the hyperbola is:
\[ \frac{x^2}{\frac{5}{2}} - \frac{y^2}{\frac{5}{3}} = 1 \]
Multiplying through by the least common multiple of the denominators (which is 6):
\[ 6 \left(\frac{x^2}{\frac{5}{2}} - \frac{y^2}{\frac{5}{3}}\right) = 6 \]
This simplifies to:
\[ \frac{12x^2}{5} - \frac{10y^2}{5} = 6 \]
Thus, the final equation is:
\[ 12x^2 - 10y^2 = 30 \quad \text{or} \quad 2x^2 - \frac{3}{5}y^2 = 1 \]