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 line at the point \((\sqrt{5}, 4)\) on the hyperbola \(4x^2 - y^2 = 4\), we can follow these steps:

### Step 1: Find the equation of the tangent line
The equation of the hyperbola is given by:
\[ 4x^2 - y^2 = 4 \]
We can rewrite this in standard form:
\[ \frac{x^2}{1} - \frac{y^2}{4} = 1 \]
Here, \(a^2 = 1\) and \(b^2 = 4\), so \(a = 1\) and \(b = 2\).

The formula for the tangent line at the point \((x_1, y_1)\) on the hyperbola is:
\[ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 \]
Substituting \(x_1 = \sqrt{5}\), \(y_1 = 4\), \(a^2 = 1\), and \(b^2 = 4\):
\[ \frac{x \cdot \sqrt{5}}{1} - \frac{y \cdot 4}{4} = 1 \]
This simplifies to:
\[ \sqrt{5} x - y = 1 \]
or rearranging gives:
\[ y = \sqrt{5} x - 1 \]

### Step 2: Find the points of intersection
Next, we need to find the points of intersection of the lines \(y = 2x\), \(y = -2x\), and \(y = \sqrt{5} x - 1\).

**Intersection of \(y = 2x\) and \(y = \sqrt{5} x - 1\)**:
Set \(2x = \sqrt{5} x - 1\):
\[ \sqrt{5} x - 2x = 1 \implies (\sqrt{5} - 2)x = 1 \implies x = \frac{1}{\sqrt{5} - 2} \]
Substituting \(x\) back into \(y = 2x\):
\[ y = 2 \cdot \frac{1}{\sqrt{5} - 2} = \frac{2}{\sqrt{5} - 2} \]
Thus, one vertex is:
\[ \left(\frac{1}{\sqrt{5} - 2}, \frac{2}{\sqrt{5} - 2}\right) \]

**Intersection of \(y = -2x\) and \(y = \sqrt{5} x - 1\)**:
Set \(-2x = \sqrt{5} x - 1\):
\[ \sqrt{5} x + 2x = 1 \implies (\sqrt{5} + 2)x = 1 \implies x = \frac{1}{\sqrt{5} + 2} \]
Substituting \(x\) back into \(y = -2x\):
\[ y = -2 \cdot \frac{1}{\sqrt{5} + 2} = -\frac{2}{\sqrt{5} + 2} \]
Thus, another vertex is:
\[ \left(\frac{1}{\sqrt{5} + 2}, -\frac{2}{\sqrt{5} + 2}\right) \]

**Intersection of \(y = 2x\) and \(y = -2x\)**:
Setting \(2x = -2x\) gives:
\[ 4x = 0 \implies x = 0 \implies y = 0 \]
Thus, the third vertex is:
\[ (0, 0) \]

### Step 3: Calculate the area of the triangle
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. \(C(0, 0)\)

Using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| \frac{1}{\sqrt{5} - 2} \left(-\frac{2}{\sqrt{5} + 2} - 0\right) + \frac{1}{\sqrt{5} + 2} \left(0 - \frac{2}{\sqrt{5} - 2}\right) + 0 \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| -\frac{2}{(\sqrt{5} - 2)(\sqrt{5} + 2)} - \frac{2}{(\sqrt{5} + 2)(\sqrt{5} - 2)} \right| \]
This results in:
\[ = \frac{1}{2} \left| -\frac{4}{1} \right| = \frac{4}{2} = 2 \text{ square units} \]

Thus, the area of the triangle is \(2\) square units.