The line `4x -7y + 10 = 0` intersects the parabola `y^(2) =4x` at the points P and Q. The coordinates of the point of intersection of the tangents drawn at the points P and Q are

Options

A. `(-(7)/(2),-(5)/(2))`
B. `((5)/(2),(7)/(2))`
C. `(-(5)/(2),-(7)/(2))`
D. `((7)/(2),(5)/(2))`

Answer

To solve the problem of finding the coordinates of the point of intersection of the tangents drawn at the points P and Q, where the line intersects the parabola, we will follow these steps:

### Step 1: Find the points of intersection of the line and the parabola.

The equations we have are:
- Line: \( 4x - 7y + 10 = 0 \)
- Parabola: \( y^2 = 4x \)

First, we can express \( x \) in terms of \( y \) from the parabola:
\[ x = \frac{y^2}{4} \]

Now, substitute this expression for \( x \) into the line equation:
\[ 4\left(\frac{y^2}{4}\right) - 7y + 10 = 0 \]
This simplifies to:
\[ y^2 - 7y + 10 = 0 \]

### Step 2: Solve the quadratic equation for \( y \).

We can factor the quadratic equation:
\[ (y - 5)(y - 2) = 0 \]
Thus, the solutions for \( y \) are:
\[ y = 5 \quad \text{and} \quad y = 2 \]

### Step 3: Find the corresponding \( x \) values for each \( y \).

Using the parabola equation \( y^2 = 4x \):
1. For \( y = 5 \):
\[ 5^2 = 4x \implies 25 = 4x \implies x = \frac{25}{4} \]
So, point \( Q \) is \( \left(\frac{25}{4}, 5\right) \).

2. For \( y = 2 \):
\[ 2^2 = 4x \implies 4 = 4x \implies x = 1 \]
So, point \( P \) is \( (1, 2) \).

### Step 4: Find the equations of the tangents at points \( P \) and \( Q \).

The general equation of the tangent to the parabola \( y^2 = 4ax \) at point \( (x_1, y_1) \) is given by:
\[ yy_1 = 2a(x + x_1) \]
Here, \( a = 1 \) (since \( 4a = 4 \)).

1. For point \( P(1, 2) \):
\[ 2y = 2(x + 1) \implies 2y = 2x + 2 \implies y = x + 1 \]

2. For point \( Q\left(\frac{25}{4}, 5\right) \):
\[ 5y = 2(x + \frac{25}{4}) \implies 5y = 2x + \frac{25}{2} \]
Rearranging gives:
\[ 5y = 2x + \frac{25}{2} \implies 10y = 4x + 25 \implies 4x - 10y + 25 = 0 \]

### Step 5: Find the intersection of the two tangent lines.

We have the two tangent equations:
1. \( y = x + 1 \) (Equation 1)
2. \( 4x - 10y + 25 = 0 \) (Equation 2)

Substituting Equation 1 into Equation 2:
\[ 4x - 10(x + 1) + 25 = 0 \]
This simplifies to:
\[ 4x - 10x - 10 + 25 = 0 \implies -6x + 15 = 0 \implies 6x = 15 \implies x = \frac{15}{6} = \frac{5}{2} \]

Now, substituting \( x = \frac{5}{2} \) back into Equation 1 to find \( y \):
\[ y = \frac{5}{2} + 1 = \frac{7}{2} \]

### Final Answer:
The coordinates of the point of intersection of the tangents drawn at points P and Q are:
\[ \left(\frac{5}{2}, \frac{7}{2}\right) \]