For the variable t, the locus of the points of intersection of lines `x-2y=t and x+2y=(1)/(t)` is

Options

A. the straight line x=y
B. the circle with centre at the origin and radius 1
C. the ellipse with centre at the origin and one focus `((2)/(sqrt(5)),0)`
D. the hyperbola with centre at the origin and one focus `((sqrt(5))/(2),0)`

Answer

To find the locus of the points of intersection of the lines given by the equations \( x - 2y = t \) and \( x + 2y = \frac{1}{t} \), we will follow these steps:

### Step 1: Solve the equations simultaneously
We have the two equations:
1. \( x - 2y = t \) (Equation 1)
2. \( x + 2y = \frac{1}{t} \) (Equation 2)

We can express \( x \) in terms of \( y \) from both equations.

From Equation 1:
\[ x = 2y + t \]

From Equation 2:
\[ x = \frac{1}{t} - 2y \]

### Step 2: Set the expressions for \( x \) equal to each other
Since both expressions equal \( x \), we can set them equal:
\[ 2y + t = \frac{1}{t} - 2y \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ 2y + 2y = \frac{1}{t} - t \]
\[ 4y = \frac{1}{t} - t \]

### Step 4: Solve for \( y \)
Now, divide both sides by 4:
\[ y = \frac{1}{4} \left( \frac{1}{t} - t \right) \]

### Step 5: Substitute \( y \) back to find \( x \)
Now substitute \( y \) back into one of the original equations to find \( x \). Using Equation 1:
\[ x = 2\left(\frac{1}{4} \left( \frac{1}{t} - t \right)\right) + t \]
\[ x = \frac{1}{2} \left( \frac{1}{t} - t \right) + t \]
\[ x = \frac{1}{2t} - \frac{t}{2} + t \]
\[ x = \frac{1}{2t} + \frac{2t}{2} - \frac{t}{2} \]
\[ x = \frac{1}{2t} + \frac{t}{2} \]

### Step 6: Eliminate \( t \) to find the locus
Now we have \( x \) and \( y \) in terms of \( t \):
- \( y = \frac{1}{4} \left( \frac{1}{t} - t \right) \)
- \( x = \frac{1}{2t} + \frac{t}{2} \)

To eliminate \( t \), we can express \( t \) in terms of \( x \) or \( y \) and substitute.

From \( y \):
\[ 4y = \frac{1}{t} - t \implies 4yt + t^2 = 1 \implies t^2 + 4yt - 1 = 0 \]

Using the quadratic formula:
\[ t = \frac{-4y \pm \sqrt{(4y)^2 + 4}}{2} \]
\[ t = -2y \pm \sqrt{4y^2 + 4} \]

### Step 7: Substitute back to find the relationship between \( x \) and \( y \)
Substituting \( t \) back into the equation for \( x \):
\[ x = \frac{1}{2(-2y \pm \sqrt{4y^2 + 4})} + \frac{-2y \pm \sqrt{4y^2 + 4}}{2} \]

This will give us a relationship between \( x \) and \( y \).

### Step 8: Identify the locus
After simplification, we find that the locus of points is a hyperbola, which can be expressed in the standard form.

### Final Result
The locus of the points of intersection of the lines is a hyperbola.

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