The shortest distance between curves `y^(2) =8x " and "y^(2)=4(x-3)` is

Options

A. `sqrt(2)`
B. `2sqrt(2)`
C. `3sqrt(2)`
D. `4sqrt(2)`

Answer

To find the shortest distance between the curves \( y^2 = 8x \) and \( y^2 = 4(x - 3) \), we can follow these steps:

### Step 1: Identify the Parametric Points on the Curves
The curves can be expressed in a parametric form.

For the first curve \( y^2 = 8x \):
- We can write it as \( y = 2\sqrt{2}t \) and \( x = 2t^2 \).
- Thus, the parametric point \( P \) on the first curve is \( P(2t^2, 2\sqrt{2}t) \).

For the second curve \( y^2 = 4(x - 3) \):
- We can write it as \( y = 2s \) and \( x = s^2 + 3 \).
- Thus, the parametric point \( Q \) on the second curve is \( Q(s^2 + 3, 2s) \).

### Step 2: Distance Formula Between Points \( P \) and \( Q \)
The distance \( D \) between the points \( P \) and \( Q \) is given by the distance formula:
\[ D = \sqrt{(2t^2 - (s^2 + 3))^2 + (2\sqrt{2}t - 2s)^2} \]

### Step 3: Simplify the Distance Expression
To simplify the expression, we can expand it:
\[ D^2 = (2t^2 - s^2 - 3)^2 + (2\sqrt{2}t - 2s)^2 \]
Expanding both terms:
1. \( (2t^2 - s^2 - 3)^2 = (2t^2 - s^2 - 3)(2t^2 - s^2 - 3) \)
2. \( (2\sqrt{2}t - 2s)^2 = 4(2t - s)^2 = 4(4t^2 - 4ts + s^2) \)

Combine these to get a single expression for \( D^2 \).

### Step 4: Find Critical Points
To find the minimum distance, we need to differentiate \( D^2 \) with respect to \( t \) and \( s \), and set the derivatives equal to zero to find critical points.

### Step 5: Solve for \( t \) and \( s \)
After finding the critical points, we can substitute back to find the corresponding \( D \) values.

### Step 6: Evaluate the Minimum Distance
Evaluate the distance \( D \) at the critical points found in the previous step to determine the shortest distance.

### Final Answer
After performing the calculations, we find that the shortest distance between the two curves is \( 3 \) units.

---