Normals at P, Q, R are drawn to `y^(2)=4x` which intersect at (3, 0). Then, area of `DeltaPQR`, is

Options

A. 4
B. 2
C. 1
D. none of these

Explanation

To find the area of triangle PQR formed by the normals to the parabola \(y^2 = 4x\) that intersect at the point (3, 0), we can follow these steps:

### Step 1: Understand the parabola and its properties
The given parabola is \(y^2 = 4x\). The vertex of this parabola is at the origin (0, 0), and it opens to the right.

### Step 2: Find the slope of the normal
The slope of the tangent to the parabola at any point \((x_1, y_1)\) is given by the derivative:
\[ \frac{dy}{dx} = \frac{2}{y} \]
Thus, the slope of the normal at that point is:
\[ m_{\text{normal}} = -\frac{1}{\text{slope of tangent}} = -\frac{y}{2} \]

### Step 3: Write the equation of the normal
The normal line at point \((x_1, y_1)\) can be expressed using the point-slope form of the line:
\[ y - y_1 = m_{\text{normal}}(x - x_1) \]
Substituting \(m_{\text{normal}} = -\frac{y_1}{2}\) and the point \((3, 0)\) where the normals intersect:
\[ y - y_1 = -\frac{y_1}{2}(x - 3) \]

### Step 4: Solve for the intersection points
Rearranging the equation gives:
\[ y = -\frac{y_1}{2}(x - 3) + y_1 \]
This can be simplified to:
\[ y = -\frac{y_1}{2}x + \frac{3y_1}{2} \]

### Step 5: Find the points of intersection with the parabola
Substituting \(y\) into the parabola's equation \(y^2 = 4x\):
\[ \left(-\frac{y_1}{2}x + \frac{3y_1}{2}\right)^2 = 4x \]
Expanding and simplifying this will yield a quadratic equation in \(x\).

### Step 6: Solve the quadratic equation
This will give us the \(x\)-coordinates of points \(P\), \(Q\), and \(R\). The corresponding \(y\)-coordinates can be found by substituting \(x\) back into the normal equations.

### Step 7: Find the coordinates of points P, Q, and R
After solving, we find that the points are:
- \(P(0, 0)\)
- \(Q(1, 2)\)
- \(R(1, -2)\)

### Step 8: Calculate the area of triangle PQR
Using the formula for the area of a triangle given by vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0(2 - (-2)) + 1((-2) - 0) + 1(0 - 2) \right| \]
\[ = \frac{1}{2} \left| 0 + (-2) + (-2) \right| = \frac{1}{2} \left| -4 \right| = \frac{4}{2} = 2 \]

### Final Answer
The area of triangle PQR is \(2\) square units.

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