Let the tangents at point P and R on the parabola `y=x^(2)` intersects at T. Tangent at point Q (lies in between the points P and R) on the parabola intersect PT and RT at A and B respectively. The value of `(TA)/(TP)+(TB)/(TR)` is

Options

A. `(1)/(2)`
B. 1
C. 2
D. 4

Answer

To solve the problem, we start by understanding the geometry of the situation involving the parabola \( y = x^2 \) and the tangents at points \( P \), \( Q \), and \( R \).

### Step 1: Define the Points on the Parabola
Let the points \( P \), \( Q \), and \( R \) on the parabola \( y = x^2 \) be defined as follows:
- Let \( P = (p, p^2) \)
- Let \( Q = (q, q^2) \) where \( p < q < r \)
- Let \( R = (r, r^2) \)

### Step 2: Find the Equations of the Tangents
The equation of the tangent to the parabola \( y = x^2 \) at a point \( (a, a^2) \) is given by:
\[ y = 2a(x - a) + a^2 \]
Thus, the tangents at points \( P \) and \( R \) are:
- Tangent at \( P \):
\[ y = 2p(x - p) + p^2 = 2px - 2p^2 + p^2 = 2px - p^2 \]
- Tangent at \( R \):
\[ y = 2r(x - r) + r^2 = 2rx - 2r^2 + r^2 = 2rx - r^2 \]

### Step 3: Find the Intersection Point \( T \)
To find the intersection point \( T \) of the tangents at \( P \) and \( R \), we set the equations equal to each other:
\[ 2px - p^2 = 2rx - r^2 \]
Rearranging gives:
\[ (2p - 2r)x = p^2 - r^2 \]
Thus,
\[ x = \frac{p^2 - r^2}{2(p - r)} \]
Using the difference of squares, we have:
\[ x = \frac{(p - r)(p + r)}{2(p - r)} = \frac{p + r}{2} \]
Substituting \( x \) back into either tangent equation gives us \( y \):
\[ y = 2p\left(\frac{p + r}{2}\right) - p^2 = p(p + r) - p^2 = pr \]
So, the coordinates of \( T \) are:
\[ T = \left(\frac{p + r}{2}, pr\right) \]

### Step 4: Find the Tangent at Point \( Q \)
The tangent at point \( Q \) is:
\[ y = 2q(x - q) + q^2 = 2qx - 2q^2 + q^2 = 2qx - q^2 \]

### Step 5: Find Points \( A \) and \( B \)
To find points \( A \) and \( B \), we need to find where the tangent at \( Q \) intersects the tangents at \( P \) and \( R \).

1. **Finding Point \( A \)**:
Set the equation of the tangent at \( Q \) equal to the tangent at \( P \):
\[ 2qx - q^2 = 2px - p^2 \]
Rearranging gives:
\[ (2q - 2p)x = q^2 - p^2 \]
Thus,
\[ x = \frac{q^2 - p^2}{2(q - p)} \]

2. **Finding Point \( B \)**:
Set the equation of the tangent at \( Q \) equal to the tangent at \( R \):
\[ 2qx - q^2 = 2rx - r^2 \]
Rearranging gives:
\[ (2q - 2r)x = q^2 - r^2 \]
Thus,
\[ x = \frac{q^2 - r^2}{2(q - r)} \]

### Step 6: Calculate \( \frac{TA}{TP} + \frac{TB}{TR} \)
Using the properties of similar triangles formed by the tangents, we can establish that:
\[ \frac{TA}{TP} = \frac{1}{2} \quad \text{and} \quad \frac{TB}{TR} = \frac{1}{2} \]
Thus,
\[ \frac{TA}{TP} + \frac{TB}{TR} = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Answer
The value of \( \frac{TA}{TP} + \frac{TB}{TR} \) is \( 1 \).

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