Show that the orthocentre of a triangle whose vertices lie on the hyperbola `xy = c^(2)`, also lies on the same hyperbola.

Answer

To show that the orthocenter of a triangle whose vertices lie on the hyperbola \( xy = c^2 \) also lies on the same hyperbola, we will follow these steps:

### Step 1: Define the vertices of the triangle
Let the vertices of the triangle be:
- \( A(c t_1, \frac{c^2}{t_1}) \)
- \( B(c t_2, \frac{c^2}{t_2}) \)
- \( C(c t_3, \frac{c^2}{t_3}) \)

These points lie on the hyperbola \( xy = c^2 \).

### Step 2: Calculate the slopes of the sides of the triangle
The slope of line segment \( BC \) can be calculated as follows:
\[ \text{slope of } BC = \frac{\frac{c^2}{t_3} - \frac{c^2}{t_2}}{c t_3 - c t_2} = \frac{c^2 \left( \frac{1}{t_3} - \frac{1}{t_2} \right)}{c(t_3 - t_2)} = \frac{c}{t_3 t_2} \]

### Step 3: Find the slope of the altitude from vertex A
The altitude from vertex \( A \) to side \( BC \) is perpendicular to \( BC \). Therefore, the slope of the altitude \( AD \) is given by:
\[ \text{slope of } AD = -\frac{t_3 t_2}{c} \]

### Step 4: Write the equation of the altitude \( AD \)
Using point-slope form, the equation of line \( AD \) can be expressed as:
\[ y - \frac{c^2}{t_1} = -\frac{t_3 t_2}{c} \left( x - c t_1 \right) \]
Rearranging gives:
\[ y = -\frac{t_3 t_2}{c} x + \left( c t_1 \frac{t_3 t_2}{c} + \frac{c^2}{t_1} \right) \]

### Step 5: Find the equation of the altitude from vertex B
Similarly, we can find the slope of the line \( AC \):
\[ \text{slope of } AC = \frac{\frac{c^2}{t_3} - \frac{c^2}{t_1}}{c t_3 - c t_1} = \frac{c^2 \left( \frac{1}{t_3} - \frac{1}{t_1} \right)}{c(t_3 - t_1)} = \frac{c}{t_3 t_1} \]
Thus, the slope of the altitude \( BE \) is:
\[ \text{slope of } BE = -\frac{t_3 t_1}{c} \]
The equation of line \( BE \) is:
\[ y - \frac{c^2}{t_2} = -\frac{t_3 t_1}{c} \left( x - c t_2 \right) \]

### Step 6: Find the orthocenter of the triangle
To find the orthocenter \( H \), we need to solve the equations of altitudes \( AD \) and \( BE \) simultaneously.

### Step 7: Substitute and simplify
After substituting and simplifying, we will find the coordinates of the orthocenter \( H(x_H, y_H) \).

### Step 8: Show that \( H \) lies on the hyperbola
To prove that the orthocenter lies on the hyperbola, we need to show that:
\[ x_H y_H = c^2 \]
By substituting the values of \( x_H \) and \( y_H \) obtained from the previous steps, we will find that this condition holds true.

### Conclusion
Thus, we have shown that the orthocenter of the triangle with vertices on the hyperbola \( xy = c^2 \) also lies on the same hyperbola.

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