The area (in sq. units) of the triangle formed by the lines `y=2x, y=-2x` and the tangent at the point `(sqrt5, 4)` on `4x^(2)-y^(2)=4` is equal to

Options

A. 4
B. 2
C. 1
D. 3

Answer

To find the area of the triangle formed by the lines \(y = 2x\), \(y = -2x\), and the tangent to the hyperbola \(4x^2 - y^2 = 4\) at the point \((\sqrt{5}, 4)\), we can follow these steps:

### Step 1: Find the equation of the tangent line at the point \((\sqrt{5}, 4)\)

The general form of the equation of the tangent to the hyperbola \(4x^2 - y^2 = 4\) at a point \((x_1, y_1)\) is given by:

\[ 4x \cdot x_1 - y \cdot y_1 = 4 \]

Substituting \(x_1 = \sqrt{5}\) and \(y_1 = 4\):

\[ 4x \cdot \sqrt{5} - y \cdot 4 = 4 \]

Simplifying this gives:

\[ 4\sqrt{5}x - 4y = 4 \]

Dividing through by 4:

\[ \sqrt{5}x - y = 1 \]

Thus, the equation of the tangent line is:

\[ y = \sqrt{5}x - 1 \]

### Step 2: Find the intersection points of the lines

#### Intersection of \(y = 2x\) and \(y = \sqrt{5}x - 1\):

Setting \(2x = \sqrt{5}x - 1\):

\[ \sqrt{5}x - 2x = 1 \]

\[ (\sqrt{5} - 2)x = 1 \]

\[ x = \frac{1}{\sqrt{5} - 2} \]

Substituting back to find \(y\):

\[ y = 2\left(\frac{1}{\sqrt{5} - 2}\right) = \frac{2}{\sqrt{5} - 2} \]

Thus, the first intersection point is:

\[ \left(\frac{1}{\sqrt{5} - 2}, \frac{2}{\sqrt{5} - 2}\right) \]

#### Intersection of \(y = -2x\) and \(y = \sqrt{5}x - 1\):

Setting \(-2x = \sqrt{5}x - 1\):

\[ \sqrt{5}x + 2x = 1 \]

\[ (\sqrt{5} + 2)x = 1 \]

\[ x = \frac{1}{\sqrt{5} + 2} \]

Substituting back to find \(y\):

\[ y = -2\left(\frac{1}{\sqrt{5} + 2}\right) = \frac{-2}{\sqrt{5} + 2} \]

Thus, the second intersection point is:

\[ \left(\frac{1}{\sqrt{5} + 2}, \frac{-2}{\sqrt{5} + 2}\right) \]

#### Intersection of \(y = 2x\) and \(y = -2x\):

Setting \(2x = -2x\):

\[ 4x = 0 \implies x = 0 \]

Thus, the third intersection point is:

\[ (0, 0) \]

### Step 3: Calculate the area of the triangle formed by these points

The vertices of the triangle are:

1. \(A\left(\frac{1}{\sqrt{5} - 2}, \frac{2}{\sqrt{5} - 2}\right)\)
2. \(B\left(\frac{1}{\sqrt{5} + 2}, \frac{-2}{\sqrt{5} + 2}\right)\)
3. \(C(0, 0)\)

The area \(A\) of the triangle can be calculated using the determinant formula:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the coordinates:

\[ \text{Area} = \frac{1}{2} \left| \frac{1}{\sqrt{5} - 2} \left(\frac{-2}{\sqrt{5} + 2} - 0\right) + \frac{1}{\sqrt{5} + 2} \left(0 - \frac{2}{\sqrt{5} - 2}\right) + 0 \right| \]

Calculating this gives:

\[ = \frac{1}{2} \left| \frac{-2}{\sqrt{5} - 2} \cdot \frac{1}{\sqrt{5} + 2} - \frac{2}{\sqrt{5} + 2} \cdot \frac{1}{\sqrt{5} - 2} \right| \]

Combining the terms:

\[ = \frac{1}{2} \left| -\frac{2}{(\sqrt{5} - 2)(\sqrt{5} + 2)} - \frac{2}{(\sqrt{5} - 2)(\sqrt{5} + 2)} \right| \]

\[ = \frac{1}{2} \left| -\frac{4}{(\sqrt{5} - 2)(\sqrt{5} + 2)} \right| \]

Calculating \((\sqrt{5} - 2)(\sqrt{5} + 2) = 5 - 4 = 1\):

\[ = \frac{1}{2} \cdot 4 = 2 \]

Thus, the area of the triangle is:

\[ \boxed{2} \]