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If P, Q, R are three points on a parabol...

If P, Q, R are three points on a parabola `y^2=4ax` whose ordinates are in geometrical progression, then the tangents at P and R meet on :

A

the line through Q parallel to the x-axis

B

the line through Q parallel to y-axis

C

the line joining Q to the vertex

D

the line joining Q to the focus

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The correct Answer is:
To solve the problem, we need to find out where the tangents at points P and R on the parabola \(y^2 = 4ax\) meet, given that the ordinates of points P, Q, and R are in geometric progression. ### Step-by-Step Solution: 1. **Identify the Points on the Parabola**: Let the coordinates of points P, Q, and R be: - \( P(t_1) = (at_1^2, 2at_1) \) - \( Q(t_2) = (at_2^2, 2at_2) \) - \( R(t_3) = (at_3^2, 2at_3) \) Since the ordinates are in geometric progression, we have: \[ 2at_1, 2at_2, 2at_3 \text{ are in GP} \] This implies: \[ (2at_2)^2 = (2at_1)(2at_3) \implies t_2^2 = t_1 t_3 \] 2. **Equation of Tangents at Points P and R**: The equation of the tangent to the parabola \(y^2 = 4ax\) at a point \((at^2, 2at)\) is given by: \[ ty = x + at^2 \] Therefore, the equations of the tangents at points P and R are: - For point P: \[ t_1y = x + at_1^2 \] - For point R: \[ t_3y = x + at_3^2 \] 3. **Finding the Intersection of the Tangents**: To find the intersection of these two tangents, we can solve the equations simultaneously: \[ t_1y - x = at_1^2 \quad \text{(1)} \] \[ t_3y - x = at_3^2 \quad \text{(2)} \] Rearranging both equations gives: \[ x = t_1y - at_1^2 \quad \text{(3)} \] \[ x = t_3y - at_3^2 \quad \text{(4)} \] Setting equations (3) and (4) equal to each other: \[ t_1y - at_1^2 = t_3y - at_3^2 \] Rearranging gives: \[ (t_1 - t_3)y = at_1^2 - at_3^2 \] \[ y = \frac{a(t_1^2 - t_3^2)}{t_1 - t_3} \] Using the difference of squares: \[ y = \frac{a(t_1 - t_3)(t_1 + t_3)}{t_1 - t_3} = a(t_1 + t_3) \] 4. **Finding the x-coordinate**: Substitute \(y = a(t_1 + t_3)\) back into equation (3): \[ x = t_1(a(t_1 + t_3)) - at_1^2 \] \[ x = at_1^2 + at_1t_3 - at_1^2 = at_1t_3 \] 5. **Conclusion**: The intersection point of the tangents at points P and R is: \[ \left(at_1t_3, a(t_1 + t_3)\right) \] Since \(t_2^2 = t_1t_3\), we can express the x-coordinate as \(at_2^2\). Thus, the intersection point lies on the line through Q, which is parallel to the y-axis. ### Final Answer: The tangents at points P and R meet on the line through Q, which is parallel to the y-axis.
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