The line 2x + y = 1 touches a hyperbola and passes through the point of intersection of a directrix and the x-axis. The equation of the hyperbola is

Options

A. `(x^(2))/(1)-(y^(2))/(3)=1`
B. `(x^(2))/(1)-(y^(2))/(3)=2`
C. `(x^(2))/(3)-(y^(2))/(1)=1`
D. `(x^(2))/(3)-(y^(2))/(1)=2`

Answer

To find the equation of the hyperbola given that the line \(2x + y = 1\) touches it and passes through the point of intersection of the directrix and the x-axis, we can follow these steps:

### Step 1: Write the equation of the line in slope-intercept form
The given line is:
\[ 2x + y = 1 \]
Rearranging it gives:
\[ y = -2x + 1 \]
Here, the slope \(m = -2\) and the y-intercept \(c = 1\).

### Step 2: Assume the standard form of the hyperbola
We assume the hyperbola is in the standard form:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 3: Use the condition for tangency
The equation of the tangent to the hyperbola at any point can be expressed as:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
Since the line touches the hyperbola, we can equate it to the tangent equation:
\[ -2x + 1 = mx + \sqrt{a^2 m^2 - b^2} \]
Here, \(m = -2\) and \(c = 1\). We can square both sides to eliminate the square root:
\[ 1 = a^2(-2)^2 - b^2 \]
This simplifies to:
\[ 1 = 4a^2 - b^2 \quad \text{(Equation 1)} \]

### Step 4: Find the point of intersection of the directrix and the x-axis
For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the directrices are given by:
\[ x = \frac{a}{e} \quad \text{and} \quad x = -\frac{a}{e} \]
where \(e = \sqrt{1 + \frac{b^2}{a^2}}\).

The intersection of the directrix with the x-axis occurs at the points \((\frac{a}{e}, 0)\) and \((- \frac{a}{e}, 0)\).

### Step 5: Use the condition that the line passes through the intersection point
The line \(2x + y = 1\) must pass through the point of intersection. We can substitute \(y = 0\) into the line equation:
\[ 2x + 0 = 1 \implies x = \frac{1}{2} \]
This means the point of intersection of the directrix and the x-axis is \(\left(\frac{1}{2}, 0\right)\).

### Step 6: Relate \(a\) and \(e\)
From the point of intersection, we have:
\[ \frac{a}{e} = \frac{1}{2} \]
Thus,
\[ a = \frac{e}{2} \]

### Step 7: Substitute \(e\) in terms of \(a\) into the equations
Using the definition of \(e\):
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \(a = \frac{e}{2}\) into the equation gives:
\[ e = \sqrt{1 + \frac{b^2}{(\frac{e}{2})^2}} = \sqrt{1 + \frac{4b^2}{e^2}} \]
Squaring both sides:
\[ e^2 = 1 + \frac{4b^2}{e^2} \]
Multiplying through by \(e^2\) gives:
\[ e^4 - e^2 - 4b^2 = 0 \]

### Step 8: Solve for \(a^2\) and \(b^2\)
Now, we have two equations:
1. \(4a^2 - b^2 = 1\)
2. \(e^4 - e^2 - 4b^2 = 0\)

Substituting \(e^2 = 1 + \frac{b^2}{a^2}\) into the second equation and solving gives us values for \(a^2\) and \(b^2\).

### Step 9: Check possible values
After solving, we find:
- If \(a^2 = 1\), then \(b^2 = 3\).
- If \(a^2 = \frac{1}{4}\), then \(b^2\) becomes negative, which is not possible.

Thus, we conclude:
\[ a^2 = 1 \quad \text{and} \quad b^2 = 3 \]

### Final Equation of the Hyperbola
The equation of the hyperbola is:
\[ \frac{x^2}{1} - \frac{y^2}{3} = 1 \quad \text{or} \quad x^2 - \frac{y^2}{3} = 1 \]

### Conclusion
The correct option for the hyperbola is:
\[ \text{Option 1: } x^2 - \frac{y^2}{3} = 1 \]