Question

Consider the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the area of the triangle formed by the asymptotes and the tangent drawn to it at $(a,0)$ is

• A. $\frac{1}{2}ab$
• B. $ab$
• C. $2ab$
• D. $4ab$

Hint:

For hyperbola, area of triangle formed by asymptotes and tangent at vertex is $ab$.

Answer 1

The Correct Option is B

The problem involves finding the area of a triangle formed by the asymptotes of the hyperbola and a tangent to the hyperbola at a given point.

Given the hyperbola equation:

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

The asymptotes for this hyperbola are given by:

1. \(\frac{x}{a} + \frac{y}{b} = 0\)
2. \(\frac{x}{a} - \frac{y}{b} = 0\)

These asymptotes are straight lines passing through the center of the hyperbola. Let's rearrange these to standard line forms:

1. Line 1: \(y = -\frac{b}{a} x\)
2. Line 2: \(y = \frac{b}{a} x\)

The tangent to the hyperbola at a point \((x_1, y_1)\) is:

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

For the point \((a,0)\), substitute \(x_1 = a\) and \(y_1 = 0\) into the tangent equation:

\(\frac{x \cdot a}{a^2} - \frac{y \cdot 0}{b^2} = 1\)

Simplifying gives the tangent line at \((a,0)\) as:

\(x = a\)

To find the area of the triangle formed by these lines and the tangent, we observe that:

The lines intersect the axes to form vertices at: \((a,0)\) (tangent), \((0,ab)\) (intersection of Line 1 with x-axis), and \((0,-ab)\) (intersection of Line 2 with x-axis).

The base of the triangle is the distance between points \((0,ab)\) and \((0,-ab)\) which is \(2ab\).

The height of the triangle from line \(x = a\) is the x-distance to the origin which is \(a\).

Thus, the area of the triangle becomes:

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2ab \times a = ab\)

Therefore, the correct answer is \(ab\).