Let a and b be positive real numbers such that ` a gt 1 and b lt a ` . Let be a point in the first quadrant that lies on the hyperbola `(x^(2))/(a^(2)) - (y^(2))/(b^(2)) = 1` . Suppose the tangent to the hyperbola at P passes through the oint (1,0) and suppose the normal to the hyperbola at P cuts off equal intercepts on the coordinate axes . Let `Delta ` denote the area of the triangle formed by the tangent at P , the normal at P and the x - axis . If a denotes the eccentricity of the hyperbola , then which of the following statements is/are TRUE ?

Options

A. `1 lt e lt sqrt(2)`
B. `sqrt(2) lt e lt 2 `
C. `Delta= a^(4)`
D. `Delta=b^(4)`

Answer

To solve the given problem, we need to analyze the hyperbola and the properties of the tangent and normal lines at a point \( P \) on the hyperbola. We will derive the necessary equations step by step.

### Step 1: Understand the hyperbola equation
The hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( a > 1 \) and \( b < a \).

### Step 2: Find the coordinates of point \( P \)
Let \( P \) be a point on the hyperbola, represented as \( P(a \sec \theta, b \tan \theta) \) for some angle \( \theta \).

### Step 3: Equation of the tangent line at point \( P \)
The equation of the tangent line at point \( P \) can be derived using the formula:
\[ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 \]
Substituting \( x_1 = a \sec \theta \) and \( y_1 = b \tan \theta \):
\[ \frac{x (a \sec \theta)}{a^2} - \frac{y (b \tan \theta)}{b^2} = 1 \]
This simplifies to:
\[ \frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1 \]

### Step 4: Condition for the tangent to pass through (1, 0)
Substituting \( (1, 0) \) into the tangent equation:
\[ \frac{1 \cdot \sec \theta}{a} - 0 = 1 \implies \sec \theta = a \]
Thus, we have:
\[ \theta = \sec^{-1}(a) \]

### Step 5: Equation of the normal line at point \( P \)
The slope of the tangent line at \( P \) is given by:
\[ m_t = \frac{b}{a} \tan \theta \]
The slope of the normal line is:
\[ m_n = -\frac{1}{m_t} = -\frac{a}{b \tan \theta} \]
Using the point-slope form, the equation of the normal line becomes:
\[ y - b \tan \theta = -\frac{a}{b \tan \theta}(x - a \sec \theta) \]

### Step 6: Normal cuts equal intercepts
For the normal to cut equal intercepts on the axes, we set the intercepts equal:
Let the intercepts be \( c \), then:
\[ \frac{c}{a} = \frac{c}{b} \implies a = b \]
This contradicts our initial condition \( b < a \). Thus, we need to find the area of the triangle formed by the tangent, normal, and x-axis.

### Step 7: Area of triangle \( \Delta \)
The area \( \Delta \) of the triangle formed by the tangent line, normal line, and x-axis can be calculated using the formula for the area of a triangle:
\[ \Delta = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the x-intercept of the normal line and the height is the y-intercept of the tangent line.

### Step 8: Find the eccentricity \( e \)
The eccentricity \( e \) of the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Given the conditions, we can analyze the range of \( e \).

### Conclusion
From the analysis, we conclude that the eccentricity \( e \) lies between 1 and \( \sqrt{2} \). Therefore, the true statement(s) regarding the eccentricity of the hyperbola can be determined based on the derived conditions.