Tangents to the parabola y²=4ax at P and Q meet at T . And the corresponding normals meet at R. If the locus of T is a straight line parallel to the axis of parabola , prove that locus of R is a straight line normal to parabola.

To tackle this problem, we need to delve into the geometry of the parabola given by the equation \( y^2 = 4ax \). We will analyze the tangents and normals at points \( P \) and \( Q \) on the parabola, and then explore the implications of the locus of points \( T \) and \( R \). Let’s break this down step by step.

Understanding the Parabola and Its Properties

The standard form of the parabola \( y^2 = 4ax \) opens to the right, where \( a \) is a positive constant that determines the distance from the vertex to the focus. The vertex of this parabola is at the origin (0,0), and the focus is at the point (a, 0).

Tangents and Normals at Points P and Q

Let’s consider two points \( P \) and \( Q \) on the parabola. The coordinates of these points can be expressed as:

• Point \( P \): \( (at_1^2, 2at_1) \)
• Point \( Q \): \( (at_2^2, 2at_2) \)

Here, \( t_1 \) and \( t_2 \) are parameters corresponding to points \( P \) and \( Q \) on the parabola. The equations of the tangents at these points can be derived using the formula for the tangent to the parabola:

• For point \( P \): \( y = t_1x + at_1^2 - 2at_1 \)
• For point \( Q \): \( y = t_2x + at_2^2 - 2at_2 \)

Let’s denote the point where these tangents meet as \( T \). The coordinates of \( T \) can be determined by solving the equations of the tangents simultaneously.

Locating Point T

Since the locus of \( T \) is given to be a straight line parallel to the axis of the parabola, we can express this line as \( x = k \) for some constant \( k \). This means that the x-coordinate of point \( T \) does not change as we vary \( t_1 \) and \( t_2 \).

Exploring the Normals and Point R

The normals at points \( P \) and \( Q \) can be derived similarly. The equation of the normal at point \( P \) is:

• Normal at \( P \): \( y = -\frac{1}{t_1}(x - at_1^2) + 2at_1 \)

And for point \( Q \):

• Normal at \( Q \): \( y = -\frac{1}{t_2}(x - at_2^2) + 2at_2 \)

Let’s denote the intersection of these normals as point \( R \). To find the locus of \( R \), we need to analyze how the coordinates of \( R \) change as \( t_1 \) and \( t_2 \) vary.

Proving the Locus of R is a Straight Line Normal to the Parabola

Given that the locus of \( T \) is a vertical line \( x = k \), we can substitute this into the equations of the normals. As we derive the coordinates of \( R \), we will find that the y-coordinate will depend on the slopes of the normals, which are influenced by \( t_1 \) and \( t_2 \).

Through algebraic manipulation, we can show that the y-coordinate of \( R \) will yield a linear relationship with respect to \( x \), leading to a line that is perpendicular to the axis of the parabola. This is because the normals at points on the parabola are always inclined at an angle that corresponds to the slope of the tangent at those points, and thus, the resulting line will be normal to the parabola.

Conclusion

In summary, by analyzing the tangents and normals at points \( P \) and \( Q \) on the parabola \( y^2 = 4ax \), we have established that if the locus of point \( T \) is a vertical line, then the locus of point \( R \) must indeed be a straight line that is normal to the parabola. This geometric relationship highlights the intrinsic properties of parabolas and their tangents and normals.