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

• A. a(1 +t2)
• B. ( 1+ t2)
• C. a( 1 - t2 )
• D. ( 1- t2)

Answer 1

The Correct Option is A

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: \[ R = a(1 + t^2) \]

Answer:

\[ \boxed{a(1 + t^2)} \]