Four points are such that the line joining any two points is perpendicular to the line joining other two points. If three point out of these lie on a rectangular hyperbola, then the fourth point will lie on

Options

A. the same hyperbola
B. the conjugate hyperbola
C. one of the directrix
D. one of the asymptotes

Explanation

To solve the problem step by step, we need to analyze the given conditions and apply the properties of a rectangular hyperbola and the orthocenter of a triangle.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have four points such that the line joining any two points is perpendicular to the line joining the other two points. This means that these points can be considered as the vertices of a cyclic quadrilateral.

**Hint**: Recall that in a cyclic quadrilateral, the opposite angles are supplementary, and the perpendicularity condition suggests a special relationship between the points.

2. **Identifying the Triangle**: Out of the four points, three points lie on a rectangular hyperbola. Let's denote these points as \( A, B, C \). The fourth point, which we will denote as \( D \), is the point we need to analyze.

**Hint**: Remember that the properties of the rectangular hyperbola will play a crucial role in determining the position of point \( D \).

3. **Orthocenter Concept**: The orthocenter of a triangle is the point where the altitudes of the triangle intersect. For triangle \( ABC \), the orthocenter is denoted as \( H \).

**Hint**: Consider the relationship between the orthocenter and the circumcircle of the triangle formed by points \( A, B, C \).

4. **Circumcircle and Rectangular Hyperbola**: A key property of a rectangular hyperbola is that it can circumscribe a triangle. If the vertices of triangle \( ABC \) lie on a rectangular hyperbola, then the orthocenter \( H \) of triangle \( ABC \) also lies on the same hyperbola.

**Hint**: Use the property that if a triangle is inscribed in a conic section (like a hyperbola), certain points related to the triangle (like the orthocenter) will also lie on that conic.

5. **Conclusion**: Since the orthocenter \( H \) of triangle \( ABC \) lies on the rectangular hyperbola, and point \( D \) is the orthocenter of triangle \( ABC \), it follows that point \( D \) must also lie on the rectangular hyperbola.

**Hint**: Conclude by stating that since all properties align, the fourth point \( D \) must lie on the same rectangular hyperbola as points \( A, B, C \).

### Final Answer:
The fourth point will lie on the same rectangular hyperbola as the other three points.