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

Let the point of contact on the hyperbola be $$P\,(x_1,y_1)$$, where $$x_1\gt 0,\;y_1\gt 0$$ and

$$\frac{x_1^{2}}{a^{2}}-\frac{y_1^{2}}{b^{2}}=1\quad -(1)$$

1. Tangent condition
For $$x^{2}/a^{2}-y^{2}/b^{2}=1$$, differentiating gives

$$\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0 \;\Longrightarrow\; \left(\frac{dy}{dx}\right)_{P}=m_t=\frac{b^{2}x_1}{a^{2}y_1}$$

The tangent at $$P$$ therefore is $$y-y_1=m_t\,(x-x_1).$$ It passes through $$(1,0)$$, hence

$$0-y_1=\frac{b^{2}x_1}{a^{2}y_1}\,(1-x_1) \;\Longrightarrow\; y_1^{2}=\frac{b^{2}x_1}{a^{2}}(x_1-1)\quad -(2)$$

2. Normal condition
The slope of the normal is $$m_n=-\frac{1}{m_t}=-\frac{a^{2}y_1}{b^{2}x_1}.$$

Because the normal cuts equal intercepts on the coordinate axes, its slope is $$-1$$. Hence

$$-\frac{a^{2}y_1}{b^{2}x_1}=-1 \;\Longrightarrow\; a^{2}y_1=b^{2}x_1 \;\Longrightarrow\; y_1=\frac{b^{2}x_1}{a^{2}}\quad -(3)$$

3. Determining $$x_1,\;y_1$$ and the relation between $$a,b$$
Substituting $$y_1$$ from $$-(3)$$ into $$-(2)$$:

$$\left(\frac{b^{2}x_1}{a^{2}}\right)^{2} =\frac{b^{2}x_1}{a^{2}}\,(x_1-1)$$

$$\frac{b^{4}x_1^{2}}{a^{4}} =\frac{b^{2}x_1}{a^{2}}\,(x_1-1)$$

$$b^{2}x_1=a^{2}(x_1-1) \;\Longrightarrow\; x_1(a^{2}-b^{2})=a^{2} \;\Longrightarrow\; x_1=\frac{a^{2}}{a^{2}-b^{2}}\quad -(4)$$

Using $$-(3)$$, $$y_1=\frac{b^{2}}{a^{2}-b^{2}}\quad -(5)$$

Now substitute $$x_1,y_1$$ back into the hyperbola equation $$-(1)$$:

$$\frac{\bigl(\tfrac{a^{2}}{a^{2}-b^{2}}\bigr)^{2}}{a^{2}} - \frac{\bigl(\tfrac{b^{2}}{a^{2}-b^{2}}\bigr)^{2}}{b^{2}} =\frac{a^{2}}{(a^{2}-b^{2})^{2}}- \frac{b^{2}}{(a^{2}-b^{2})^{2}} =\frac{a^{2}-b^{2}}{(a^{2}-b^{2})^{2}} =\frac{1}{a^{2}-b^{2}}=1$$

Therefore $$a^{2}-b^{2}=1\quad -(6)$$

With $$-(6)$$, formulas $$-(4),(5)$$ simplify to

$$x_1=a^{2},\qquad y_1=b^{2}$$

4. The eccentricity
For $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$, the eccentricity is

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

Because $$a\gt 1$$, $$\dfrac{1}{a^{2}} \in (0,1)$$, so

$$1\lt e\lt\sqrt{2}$$

This verifies statement A and renders statement B false.

5. Area of the triangle
• Tangent meets the $$x$$-axis at $$(1,0)$$.
• Normal has equation $$x+y=k$$ (equal intercepts), passes through $$P(a^{2},b^{2})$$, hence $$k=a^{2}+b^{2}$$.
  ⇒ Normal meets the $$x$$-axis at $$(k,0)\,=(a^{2}+b^{2},0)$$.

The triangle’s vertices are $$(1,0),\;(a^{2}+b^{2},0),\;(a^{2},b^{2})$$.

Base on the $$x$$-axis  =  $$(a^{2}+b^{2})-1 = a^{2}+b^{2}-1$$.
Using $$b^{2}=a^{2}-1$$ from $$-(6)$$, $$a^{2}+b^{2}-1 = a^{2}+a^{2}-1-1 = 2(a^{2}-1).$$

Height  =  $$y_1=b^{2}=a^{2}-1.$$ Therefore

$$\Delta=\tfrac12\bigl[2(a^{2}-1)\bigr](a^{2}-1)=(a^{2}-1)^{2}=b^{4}$$

So statement D is true, while statement C ($$\Delta=a^{4}$$) is false.

6. Final verdict
The true statements are:
Option A: $$1\lt e\lt\sqrt{2}$$
Option D: $$\Delta=b^{4}$$

Hence the correct answer is:
Option A and Option D.