If A is the area of triangle formed by positive x-axis and the normal and the tangents to `x^(2) + y^(2) = 9` at `(1, sqrt8)` then A is equal to `(sqrt2= 1.41)`

Answer

To find the area \( A \) of the triangle formed by the positive x-axis, the normal, and the tangent to the curve \( x^2 + y^2 = 9 \) at the point \( (1, \sqrt{8}) \), we will follow these steps:

### Step 1: Find the slope of the tangent line at the point \( (1, \sqrt{8}) \).

The equation of the circle is given by:
\[ x^2 + y^2 = 9 \]

Differentiating both sides with respect to \( x \):
\[ 2x + 2y \frac{dy}{dx} = 0 \]

Rearranging gives:
\[ \frac{dy}{dx} = -\frac{x}{y} \]

Now, substituting the point \( (1, \sqrt{8}) \):
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{8}} = -\frac{1}{2\sqrt{2}} \]

### Step 2: Find the equation of the tangent line.

Using the point-slope form of the line:
\[ y - y_1 = m (x - x_1) \]
where \( (x_1, y_1) = (1, \sqrt{8}) \) and \( m = -\frac{1}{2\sqrt{2}} \):
\[ y - \sqrt{8} = -\frac{1}{2\sqrt{2}}(x - 1) \]

Rearranging this equation:
\[ y = -\frac{1}{2\sqrt{2}}x + \frac{1}{2\sqrt{2}} + \sqrt{8} \]
\[ y = -\frac{1}{2\sqrt{2}}x + \frac{1}{2\sqrt{2}} + 2\sqrt{2} \]
\[ y = -\frac{1}{2\sqrt{2}}x + \frac{5}{2\sqrt{2}} \]

### Step 3: Find the slope of the normal line.

The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ m_{normal} = -\frac{1}{-\frac{1}{2\sqrt{2}}} = 2\sqrt{2} \]

### Step 4: Find the equation of the normal line.

Using the point-slope form again:
\[ y - \sqrt{8} = 2\sqrt{2}(x - 1) \]
Rearranging gives:
\[ y = 2\sqrt{2}x - 2\sqrt{2} + \sqrt{8} \]
\[ y = 2\sqrt{2}x - 2\sqrt{2} + 2\sqrt{2} \]
\[ y = 2\sqrt{2}x \]

### Step 5: Find the points where the tangent and normal intersect the x-axis.

For the tangent line, set \( y = 0 \):
\[ 0 = -\frac{1}{2\sqrt{2}}x + \frac{5}{2\sqrt{2}} \]
\[ \frac{1}{2\sqrt{2}}x = \frac{5}{2\sqrt{2}} \implies x = 5 \]
So the tangent intersects the x-axis at \( (5, 0) \).

For the normal line, set \( y = 0 \):
\[ 0 = 2\sqrt{2}x \implies x = 0 \]
So the normal intersects the x-axis at \( (0, 0) \).

### Step 6: Find the area of the triangle formed by these points and the origin.

The vertices of the triangle are \( (0, 0) \), \( (5, 0) \), and the intersection of the tangent and normal lines. The height of the triangle is the y-coordinate of the point where the tangent and normal intersect.

To find the area \( A \):
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
The base is the distance along the x-axis from \( (0, 0) \) to \( (5, 0) \), which is \( 5 \).

The height is the y-coordinate at \( x = 1 \) from the tangent line:
\[ y = -\frac{1}{2\sqrt{2}}(1) + \frac{5}{2\sqrt{2}} = \frac{4}{2\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2} \]

Thus, the area is:
\[ A = \frac{1}{2} \times 5 \times \sqrt{2} = \frac{5\sqrt{2}}{2} \]

### Final Result
Using the approximation \( \sqrt{2} \approx 1.41 \):
\[ A \approx \frac{5 \times 1.41}{2} \approx 3.525 \]

However, based on the video transcript, the area is stated to be \( 1.41 \). This discrepancy suggests a reevaluation of the area calculation or the interpretation of the triangle's vertices.