Let a and b be positive real numbers such that a 1 and b

Q: Wait, why is the slope of the normal $-1$ if it cuts off equal intercepts?

Great question! If a line cuts off equal intercepts on the axes, its equation is $\frac{x}{k} + \frac{y}{k} = 1$, which simplifies to $x + y = k$. The slope of this line is $-1$. Since our point $P$ is in the first quadrant and the hyperbola is opening right, the normal must have a negative slope to intersect both axes in the first quadrant. Thus, the slope is indeed $-1$.

Q: How did we get $x_1 = a^2$ so quickly?

The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $P(x_1, y_1)$ is $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$. Since this tangent passes through $(1, 0)$, we substitute $x=1$ and $y=0$ into the equation. This gives $\frac{1 \cdot x_1}{a^2} - 0 = 1$, which directly simplifies to $x_1 = a^2$. It is a beautiful, direct consequence of the geometry!

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure coordinate geometry problem, not a physics problem! While physics often uses energy conservation, here we are dealing with the algebraic properties of conic sections. The 'energy' here is the consistency of the geometric constraints—the tangent and normal must satisfy the hyperbola's equation and their own perpendicularity. Stick to the coordinate geometry toolkit; it is the most powerful weapon for this problem.

Question

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 $\Delta$ 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?

Options:

(A) (A) $1 < e < \sqrt{2}$

(B) (B) $\sqrt{2} < e < 2$

(C) (C) $\Delta = a^4$

(D) (D) $\Delta = b^4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the first quadrant of the Cartesian plane, looking at a sweeping hyperbola defined by:

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

It is a curve of infinite grace, and today, we are going to dissect its behavior at a specific point $P(x_1, y_1)$. This point is the anchor for a tangent line that passes through the coordinate $(1, 0)$ and a normal line that displays a fascinating symmetry by cutting off equal intercepts on the axes.

### The Tangent's Secret

We start by invoking the standard equation of the tangent to the hyperbola at $P(x_1, y_1)$, which is given by:

$$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$$

The problem states that this line passes through $(1, 0)$. When we substitute these coordinates into our tangent equation, the $y$-term vanishes, leaving us with:

$$\frac{x_1}{a^2} = 1 \Rightarrow x_1 = a^2$$

### The Normal's Symmetry

Now, consider the normal at $P$. The normal is the line perpendicular to the tangent, and we are told it cuts off equal intercepts on the axes. A line with equal intercepts $k$ has the equation $x + y = k$, which implies a slope of $-1$.

Since the tangent and normal are perpendicular, the product of their slopes must be $-1$. If the normal has a slope of $-1$, the tangent must have a slope of $m_T = 1$.

We can also find the slope of the tangent by differentiating the hyperbola's equation:

$$\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{b^2 x}{a^2 y}$$

At point $P$, this slope is $\frac{b^2 x_1}{a^2 y_1}$. Equating this to $1$ and substituting $x_1 = a^2$, the $a^2$ terms cancel out:

$$\frac{b^2 a^2}{a^2 y_1} = 1 \Rightarrow y_1 = b^2$$

### The Fundamental Relationship

We now know $P$ is at $(a^2, b^2)$. Since $P$ lies on the hyperbola, it must satisfy:

$$\frac{(a^2)^2}{a^2} - \frac{(b^2)^2}{b^2} = 1 \Rightarrow a^2 - b^2 = 1$$

This is the hidden key to the entire problem, defining the relationship between the hyperbola's parameters $a$ and $b$. With this, we can find the eccentricity $e$:

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

Since $a > 1$, we know $0 < \frac{1}{a^2} < 1$, which implies $1 < e^2 < 2$, or **$1 < e < \sqrt{2}$**.

### The Area of the Triangle

Finally, let us calculate the area $\Delta$ of the triangle formed by the tangent, the normal, and the $x$-axis. The vertices are $P(a^2, b^2)$, $A(1, 0)$, and the $x$-intercept of the normal.

The normal passes through $P(a^2, b^2)$ with slope $-1$, so its equation is $y - b^2 = -1(x - a^2)$. Setting $y=0$ to find the $x$-intercept $B$:

$$-b^2 = -x + a^2 \Rightarrow x = a^2 + b^2$$

The base of the triangle is the distance between $A(1, 0)$ and $B(a^2 + b^2, 0)$, which is $(a^2 + b^2) - 1$. Using $a^2 - 1 = b^2$, the base becomes $b^2 + b^2 = 2b^2$.

The height is the $y$-coordinate of $P$, which is $b^2$. The area $\Delta$ is:

$$\Delta = \frac{1}{2} 	imes 	ext{Base} 	imes 	ext{Height} = \frac{1}{2} 	imes (2b^2) 	imes (b^2) = b^4$$

Thus, the area of the triangle is **$b^4$**.

Select Answer:

Visualized Solution

Point $P$ on Hyperbola

Tangent at $P$ passing through $(1, 0)$

Finding $y_{1}$

Normal and its Intercepts

Slope of Tangent

Solving for $y_{1}$

Relation between $a$ and $b$

Eccentricity of Hyperbola

Range of Eccentricity $e$

Triangle Formed by Tangent and Normal

Finding Point $B$

Base and Height of Triangle

Area of Triangle $Δ$

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent's Secret

The Normal's Symmetry

The Fundamental Relationship

The Area of the Triangle

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