Hyperbola Tangent Normal Intercepts Area and Eccentricity

To determine which of the given statements are true, we analyze the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a > 1$ and $b < a$.

Step 1: Coordinates of Point $P$

Let $P(a \sec \theta, b \tan \theta)$ be a point on the hyperbola located in the first quadrant, so $\theta \in \left(0, \frac{\pi}{2}\right)$.

The equation of the tangent to the hyperbola at $P$ is given by: $\frac{x \sec \theta}{a} - \frac{y \tan \theta}{b} = 1$

Since the tangent passes through the point $(1, 0)$, we substitute $(1, 0)$ into the tangent equation: $\frac{1 \cdot \sec \theta}{a} - 0 = 1 \implies \sec \theta = a$

Since $\sec \theta = a > 1$, we get: $\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{a^2 - 1}$

Thus, the coordinates of $P$ are: $P = \left(a^2, b\sqrt{a^2 - 1}\right)$

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Step 2: Equation of the Normal and Finding the Relation Between $a$ and $b$

The equation of the normal to the hyperbola at $P(a \sec \theta, b \tan \theta)$ is: $\frac{a x}{\sec \theta} + \frac{b y}{\tan \theta} = a^2 + b^2$

Substituting $\sec \theta = a$ and $\tan \theta = \sqrt{a^2 - 1}$: $\frac{a x}{a} + \frac{b y}{\sqrt{a^2 - 1}} = a^2 + b^2 \implies x + \left(\frac{b}{\sqrt{a^2 - 1}}\right) y = a^2 + b^2$

The problem states that the normal cuts off equal intercepts on the coordinate axes. The $x$-intercept is $a^2 + b^2$ and the $y$-intercept is $\frac{\sqrt{a^2 - 1}}{b}(a^2 + b^2)$. Equating the magnitudes of the two intercepts gives: $\frac{b}{\sqrt{a^2 - 1}} = 1 \implies b = \sqrt{a^2 - 1} \implies b^2 = a^2 - 1$

Notice that $b = \sqrt{a^2 - 1} < a$ for all $a > 1$, which is consistent with the given condition $b < a$.

Also, substituting $b = \sqrt{a^2 - 1}$ into the coordinates of $P$, we get: $P = (a^2, b^2)$

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Step 3: Determining the Eccentricity $e$

The eccentricity $e$ of the hyperbola is: $e = \sqrt{1 + \frac{b^2}{a^2}}$

Using $b^2 = a^2 - 1$: $e = \sqrt{1 + \frac{a^2 - 1}{a^2}} = \sqrt{2 - \frac{1}{a^2}}$

Since $a > 1$, it follows that: $0 < \frac{1}{a^2} < 1 \implies 1 < 2 - \frac{1}{a^2} < 2$

Taking the square root: $1 < e < \sqrt{2}$

Thus, Option (A) is correct and Option (B) is incorrect.

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Step 4: Area of Triangle $\Delta$

The triangle is bounded by the tangent at $P$, the normal at $P$, and the $x$-axis.

1. $x$-intercept of the tangent: Setting $y = 0$ in the tangent equation gives the $x$-intercept at $x_T = 1$.

2. $x$-intercept of the normal: Setting $y = 0$ in the normal equation $x + y = a^2 + b^2$ gives the $x$-intercept at $x_N = a^2 + b^2$.

3. Length of the base on the $x$-axis: $\text{Base} = |x_N - x_T| = (a^2 + b^2) - 1$ Substituting $a^2 = b^2 + 1$: $\text{Base} = (b^2 + 1 + b^2) - 1 = 2b^2$

4. Height of the triangle: The height is the $y$-coordinate of point $P$, which is $y_P = b^2$.

5. Area of the triangle $\Delta$: $\Delta = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (2b^2) \times b^2 = b^4$

Thus, Option (D) is correct and Option (C) is incorrect.

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Conclusion

• (A) $1 < e < \sqrt{2}$ is TRUE.
• (B) $\sqrt{2} < e < 2$ is FALSE.
• (C) $\Delta = a^4$ is FALSE.
• (D) $\Delta = b^4$ is TRUE.

Correct Options: (A) and (D)