The point of intersection of the tangents at the ends of the latus rectum of the prabola `y^2 = 4x` is

Options

A. `(-1,-1)`
B. `(0, - 1)`
C. `(-1,0)`
D. `(1, 1)`

Answer

To find the point of intersection of the tangents at the ends of the latus rectum of the parabola \( y^2 = 4x \), we can follow these steps:

### Step 1: Identify the ends of the latus rectum
The latus rectum of the parabola \( y^2 = 4x \) is a line segment perpendicular to the axis of symmetry of the parabola and passes through the focus. The focus of the parabola is at \( (1, 0) \). The endpoints of the latus rectum can be found by substituting \( x = 1 \) into the equation of the parabola:

\[ y^2 = 4(1) \implies y^2 = 4 \implies y = \pm 2 \]

Thus, the endpoints of the latus rectum are \( (1, 2) \) and \( (1, -2) \).

### Step 2: Write the equations of the tangents at the endpoints
The general equation of the tangent to the parabola \( y^2 = 4x \) at a point \( (x_1, y_1) \) is given by:

\[ yy_1 = 2(x + x_1) \]

#### For the point \( (1, 2) \):
Substituting \( x_1 = 1 \) and \( y_1 = 2 \):

\[ y \cdot 2 = 2(x + 1) \implies 2y = 2x + 2 \implies x - y + 1 = 0 \quad \text{(Equation 1)} \]

#### For the point \( (1, -2) \):
Substituting \( x_1 = 1 \) and \( y_1 = -2 \):

\[ y \cdot (-2) = 2(x + 1) \implies -2y = 2x + 2 \implies 2x + 2y + 2 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( x - y + 1 = 0 \)
2. \( 2x + 2y + 2 = 0 \)

We can rewrite Equation 1 as:

\[ x = y - 1 \]

Substituting this into Equation 2:

\[ 2(y - 1) + 2y + 2 = 0 \]
\[ 2y - 2 + 2y + 2 = 0 \]
\[ 4y = 0 \implies y = 0 \]

Now substituting \( y = 0 \) back into Equation 1:

\[ x - 0 + 1 = 0 \implies x = -1 \]

### Step 4: Conclusion
Thus, the point of intersection of the tangents at the ends of the latus rectum is:

\[ \boxed{(-1, 0)} \]