Question:

Let the tangent and normal at any point $P(at^2, 2at), (a>0)$, on the parabola $y^2 = 4ax$ meet the axis of the parabola at $T$ and $G$ respectively. Then the radius of the circle through $P$, $T$, and $G$ is

Updated On: Jan 29, 2026

a(1 +t2)
( 1+ t2)
a( 1 - t2 )
( 1- t2)

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The Correct Option is A

Solution and Explanation

We are given the parabola $ y^2 = 4ax $ and the point $ P(at^2, 2at) $, where $ t $ is the parameter. The tangent and normal at any point on the parabola intersect the x-axis at the points $ T $ and $ G $, respectively.

Step 1: Equation of the tangent
The general equation of the tangent at any point $ P(x_1, y_1) $ on the parabola $ y^2 = 4ax $ is given by: \[ yy_1 = 2a(x + x_1) \] For the point $ P(at^2, 2at) $, the equation of the tangent becomes: \[ y(2at) = 2a(x + at^2) \] Simplifying, we get the equation of the tangent as: \[ y = t(x + at^2) \] This is the equation of the tangent.

Step 2: Finding the point $ T $ on the x-axis
To find the x-intercept of the tangent (point $ T $), set $ y = 0 $ in the tangent equation: \[ 0 = t(x + at^2) \] Solving for $ x $, we get: \[ x = -at^2 \] So, the point $ T $ is $ (-at^2, 0) $.

Step 3: Equation of the normal
The equation of the normal at any point $ P(x_1, y_1) $ on the parabola is: \[ y_1(y - y_1) = 2a(x - x_1) \] For the point $ P(at^2, 2at) $, the equation of the normal becomes: \[ 2at(y - 2at) = 2a(x - at^2) \] Simplifying, we get the equation of the normal: \[ y = -\frac{1}{t}(x - at^2) + 2at \] This is the equation of the normal.

Step 4: Finding the point $ G $ on the x-axis
To find the x-intercept of the normal (point $ G $), set $ y = 0 $ in the normal equation: \[ 0 = -\frac{1}{t}(x - at^2) + 2at \] Solving for $ x $, we get: \[ x = a(1 - t^2) \] So, the point $ G $ is $ (a(1 - t^2), 0) $.

Step 5: Finding the radius of the circle through $ P $, $ T $, and $ G $
The points $ P $, $ T $, and $ G $ form a circle. The radius $ R $ of the circle through these points is given by the formula: \[ R = \frac{1}{2} \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the coordinates of $ P $, $ T $, and $ G $, we get: \[ R = a(1 + t^2) \]