If chord BC subtends right angle at the vertex A of the parabola `y^(2)=4x` with `AB=sqrt(5)` then find the area of triangle ABC.

Explanation

To find the area of triangle ABC where chord BC subtends a right angle at the vertex A of the parabola \( y^2 = 4x \) and \( AB = \sqrt{5} \), we can follow these steps:

### Step 1: Identify the vertex and point B
The vertex A of the parabola \( y^2 = 4x \) is at the origin, \( A(0, 0) \).

Let point B on the parabola be represented in terms of a parameter \( t \):
- The coordinates of point B can be expressed as \( B(t) = (t^2, 2t) \).

### Step 2: Use the distance formula to find \( t \)
We know that the distance \( AB \) is given as \( \sqrt{5} \). Using the distance formula:
\[ AB = \sqrt{(t^2 - 0)^2 + (2t - 0)^2} = \sqrt{t^4 + 4t^2} \]
Setting this equal to \( \sqrt{5} \):
\[ \sqrt{t^4 + 4t^2} = \sqrt{5} \]
Squaring both sides gives:
\[ t^4 + 4t^2 = 5 \]
Rearranging this equation:
\[ t^4 + 4t^2 - 5 = 0 \]

### Step 3: Solve the quadratic in \( t^2 \)
Let \( u = t^2 \). The equation becomes:
\[ u^2 + 4u - 5 = 0 \]
Using the quadratic formula:
\[ u = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} \]
\[ u = \frac{-4 \pm 6}{2} \]
This gives:
\[ u = 1 \quad \text{or} \quad u = -5 \]
Since \( u = t^2 \) must be non-negative, we have \( t^2 = 1 \) which implies \( t = 1 \) or \( t = -1 \).

### Step 4: Find coordinates of point B
For \( t = 1 \):
\[ B(1) = (1^2, 2 \cdot 1) = (1, 2) \]
For \( t = -1 \):
\[ B(-1) = ((-1)^2, 2 \cdot (-1)) = (1, -2) \]
Thus, we can take \( B(1, 2) \) for our calculations.

### Step 5: Find coordinates of point C
Since chord BC subtends a right angle at A, we have:
\[ t \cdot t' = -4 \]
If \( t = 1 \), then:
\[ 1 \cdot t' = -4 \implies t' = -4 \]
The coordinates of point C will be:
\[ C(t') = (t'^2, 2t') = ((-4)^2, 2 \cdot (-4)) = (16, -8) \]

### Step 6: Calculate the area of triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( AB = \sqrt{5} \) and the height is the vertical distance from A to line BC.

First, we find the length of AC:
\[ AC = \sqrt{(16 - 0)^2 + (-8 - 0)^2} = \sqrt{16^2 + (-8)^2} = \sqrt{256 + 64} = \sqrt{320} \]

Now, substituting into the area formula:
\[ \text{Area} = \frac{1}{2} \times \sqrt{5} \times \sqrt{320} = \frac{1}{2} \times \sqrt{5 \times 320} = \frac{1}{2} \times \sqrt{1600} = \frac{40}{2} = 20 \]

### Final Answer
The area of triangle ABC is \( 20 \) square units.

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