Let the line y = log (a) (a > 0, a ¹ 1) intersects the line y = x at point P. If point P lies on parabola
y2 = x, then focal distance of point P is

To solve the problem, we need to find the coordinates of point P where the line \( y = \log_a(x) \) intersects the line \( y = x \), and then determine the focal distance of point P from the focus of the parabola \( y^2 = x \).

Finding the Intersection Point P

First, we set the two equations equal to each other to find the intersection point:

Since \( y = x \), we can substitute \( x \) into the logarithmic equation:

Let \( y = x \), then:

• \( x = \log_a(x) \)

To solve this equation, we can rewrite the logarithmic equation in exponential form:

Using the property of logarithms, we have:

• \( a^x = x \)

This equation does not have a straightforward algebraic solution, but we can analyze it graphically or numerically. The function \( f(x) = a^x - x \) will help us find the intersection. We know that for \( a > 1 \), \( a^x \) grows faster than \( x \) for large \( x \), and for \( 0 < a < 1 \), \( a^x \) decreases while \( x \) increases. Thus, there will be one intersection point in both cases.

Coordinates of Point P

Let’s denote the intersection point as \( P(x_0, y_0) \). Since \( y_0 = x_0 \), we can say:

At point P, we have:

• \( P(x_0, x_0) \)

Verifying Point P on the Parabola

Next, we need to confirm that point P lies on the parabola defined by \( y^2 = x \). Substituting \( y_0 = x_0 \) into the parabola's equation gives:

Substituting \( y_0 \):

• \( (x_0)^2 = x_0 \)

This simplifies to:

• \( x_0^2 - x_0 = 0 \)

Factoring out \( x_0 \), we find:

• \( x_0(x_0 - 1) = 0 \)

This yields two solutions: \( x_0 = 0 \) or \( x_0 = 1 \). Since \( a > 0 \) and \( a \neq 1 \), the relevant solution is \( x_0 = 1 \). Thus, point P is \( (1, 1) \).

Calculating the Focal Distance

The parabola \( y^2 = x \) has its focus at the point \( (1/4, 0) \). The focal distance is the distance from point P to the focus of the parabola. We can calculate this distance using the distance formula:

The distance \( d \) from point \( P(1, 1) \) to the focus \( F(1/4, 0) \) is given by:

Using the distance formula:

• \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Substituting the coordinates:

• \( d = \sqrt{(1 - 1/4)^2 + (1 - 0)^2} \)

This simplifies to:

• \( d = \sqrt{(3/4)^2 + 1^2} \)

Calculating further:

• \( d = \sqrt{9/16 + 16/16} = \sqrt{25/16} = \frac{5}{4} \)

Final Result

Thus, the focal distance of point P from the focus of the parabola \( y^2 = x \) is \( \frac{5}{4} \).