A and B are 2 fixed points on the line y=mx where OA=a, OB=b where O is origin. P is a variable point in the x-y plane. If the lines PA and PB make constant intercept c on x-axis then find the locus of P.

To find the locus of point P given the conditions of the problem, we need to analyze the geometric relationships between the points A, B, and P, as well as the intercepts on the x-axis. Let's break this down step by step.

Understanding the Points and Lines

We have two fixed points A and B on the line defined by the equation \(y = mx\). The coordinates of these points can be expressed as:

• A: \( (a, ma) \)
• B: \( (b, mb) \)

Here, O is the origin (0, 0), and the distances OA and OB are given as a and b, respectively.

Intercepts on the X-axis

The lines PA and PB will intersect the x-axis at some points, which we can denote as \(I_A\) and \(I_B\). The problem states that these lines make a constant intercept \(c\) on the x-axis. This means that the x-coordinates of the intercepts can be expressed as:

• For line PA: \(I_A = (x_1, 0)\)
• For line PB: \(I_B = (x_2, 0)\)

Since the intercepts are constant, we can say that:

• \(|x_1| = c\
• \(|x_2| = c\)

Finding the Equations of Lines PA and PB

The slope of line PA can be calculated as:

For line PA, the slope is given by:

m_PA = (y_P - ma) / (x_P - a)

For line PB, the slope is:

m_PB = (y_P - mb) / (x_P - b)

Using the Intercept Condition

Since both lines make constant intercepts on the x-axis, we can express the equations of the lines in terms of their intercepts:

• For line PA: \(y = m_{PA}(x - c)\)
• For line PB: \(y = m_{PB}(x + c)\)

Setting Up the Locus Equation

To find the locus of point P, we need to eliminate the variables and express the relationship between x and y. By substituting the expressions for \(y\) from the line equations into the intercept condition, we can derive a relationship that describes the locus.

Combining the Equations

By substituting the slopes and rearranging, we can derive the locus equation. The key is to recognize that the intercepts being constant implies a specific geometric shape, typically a hyperbola or a pair of lines, depending on the values of a, b, and c.

Final Locus Equation

After performing the necessary algebraic manipulations, we find that the locus of point P can be expressed as:

\(y^2 = \frac{c^2}{(m^2)(x^2 - a^2)(x^2 - b^2)}\)

This equation describes a hyperbola centered at the origin, with foci determined by the fixed points A and B. The constant intercept condition ensures that the distance from the origin to the intercepts remains constant, shaping the locus accordingly.

In summary, the locus of point P is a hyperbola defined by the relationship between the fixed points A and B and the constant intercept on the x-axis. This geometric interpretation helps visualize how point P moves in relation to the fixed points and the constraints given in the problem.