Minimum distance between` y^2-4x - 8y +40 =0` and `x^2 - 8x-4y +40=0` is

Options

A. 0
B. `sqrt(3)`
C. `2sqrt(2)`
D. `sqrt(2)`

Answer

To find the minimum distance between the curves given by the equations \( y^2 - 4x - 8y + 40 = 0 \) and \( x^2 - 8x - 4y + 40 = 0 \), we will follow these steps:

### Step 1: Rewrite the equations in standard form
First, we can rewrite both equations to identify their forms.

1. For the first equation:
\[ y^2 - 8y + 40 - 4x = 0 \implies y^2 - 8y + (40 - 4x) = 0 \]
This is a quadratic in \( y \).

2. For the second equation:
\[ x^2 - 8x + 40 - 4y = 0 \implies x^2 - 8x + (40 - 4y) = 0 \]
This is a quadratic in \( x \).

### Step 2: Find the slopes of the tangents
Next, we differentiate both equations to find the slopes of the tangents.

1. Differentiate the first equation with respect to \( x \):
\[ 2y \frac{dy}{dx} - 4 - 8 \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ (2y - 8) \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y - 8} \]

2. Differentiate the second equation with respect to \( y \):
\[ 2x - 8 - 4 \frac{dx}{dy} = 0 \]
Rearranging gives:
\[ 4 \frac{dx}{dy} = 2x - 8 \implies \frac{dx}{dy} = \frac{2x - 8}{4} = \frac{x - 4}{2} \]

### Step 3: Find points of common tangents
Since the curves are symmetric about the line \( y = x \), the points of tangency will be related. We can set the slopes equal to find the points where the tangents are common.

Setting \( \frac{dy}{dx} = 1 \) (since they are symmetric):
\[ \frac{4}{2y - 8} = 1 \implies 4 = 2y - 8 \implies 2y = 12 \implies y = 6 \]

### Step 4: Find corresponding \( x \) values
Substituting \( y = 6 \) back into the first equation to find \( x \):
\[ 6^2 - 4x - 8(6) + 40 = 0 \implies 36 - 4x - 48 + 40 = 0 \implies 28 - 4x = 0 \implies 4x = 28 \implies x = 7 \]
Thus, the point \( P \) on the first curve is \( (7, 6) \).

For the second curve, since they are symmetric about \( y = x \), the corresponding point \( Q \) will be \( (6, 7) \).

### Step 5: Calculate the distance between points \( P \) and \( Q \)
Now we calculate the distance \( d \) between points \( P(7, 6) \) and \( Q(6, 7) \):
\[ d = \sqrt{(7 - 6)^2 + (6 - 7)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Conclusion
The minimum distance between the two curves is:
\[ \boxed{\sqrt{2}} \]