The parabola `y^2 = kx` makes an intercept of length 4 on the line `x-2y=1` then k is

Options

A. `1`
B. `-1`
C. `2`
D. `-2`

Answer

To solve the problem, we need to find the value of \( k \) for the parabola \( y^2 = kx \) that makes an intercept of length 4 on the line \( x - 2y = 1 \).

### Step 1: Find the points of intersection
We start by substituting \( y^2 = kx \) into the line equation \( x - 2y = 1 \).

From the line equation, we can express \( x \) in terms of \( y \):
\[ x = 2y + 1 \]

Now, substitute this expression for \( x \) into the parabola equation:
\[ y^2 = k(2y + 1) \]
This simplifies to:
\[ y^2 = 2ky + k \]
Rearranging gives us:
\[ y^2 - 2ky - k = 0 \]

### Step 2: Use the quadratic formula
This is a quadratic equation in \( y \). The roots of this equation, \( y_1 \) and \( y_2 \), can be found using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2k \), and \( c = -k \):
\[ y = \frac{2k \pm \sqrt{(-2k)^2 - 4 \cdot 1 \cdot (-k)}}{2 \cdot 1} \]
This simplifies to:
\[ y = \frac{2k \pm \sqrt{4k^2 + 4k}}{2} \]
\[ y = k \pm \sqrt{k^2 + k} \]

### Step 3: Find the corresponding \( x \) values
Now, we find the corresponding \( x \) values using \( x = 2y + 1 \):
For \( y_1 = k + \sqrt{k^2 + k} \):
\[ x_1 = 2(k + \sqrt{k^2 + k}) + 1 = 2k + 2\sqrt{k^2 + k} + 1 \]

For \( y_2 = k - \sqrt{k^2 + k} \):
\[ x_2 = 2(k - \sqrt{k^2 + k}) + 1 = 2k - 2\sqrt{k^2 + k} + 1 \]

### Step 4: Calculate the length of the intercept
The length of the intercept on the line is given by the distance between the two points \( (x_1, y_1) \) and \( (x_2, y_2) \):
\[ \text{Length} = |x_1 - x_2| \]
Calculating \( x_1 - x_2 \):
\[ x_1 - x_2 = [2k + 2\sqrt{k^2 + k} + 1] - [2k - 2\sqrt{k^2 + k} + 1] \]
This simplifies to:
\[ x_1 - x_2 = 4\sqrt{k^2 + k} \]

### Step 5: Set the length equal to 4
Given that the length of the intercept is 4, we set up the equation:
\[ 4\sqrt{k^2 + k} = 4 \]
Dividing both sides by 4:
\[ \sqrt{k^2 + k} = 1 \]
Squaring both sides gives:
\[ k^2 + k = 1 \]
Rearranging this gives us:
\[ k^2 + k - 1 = 0 \]

### Step 6: Solve the quadratic equation
Now we can solve for \( k \) using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \):
\[ k = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ k = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ k = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 7: Determine the valid value of \( k \)
Since \( k \) must be non-negative for the parabola to open to the right, we take:
\[ k = \frac{-1 + \sqrt{5}}{2} \]

Thus, the value of \( k \) is:
\[ \boxed{\frac{-1 + \sqrt{5}}{2}} \]