To solve the problem, we need to find the shortest distance between the hyperbola defined by the equation \(xy = 4\) and the line given by \(y + 2x = 4\). Here’s a step-by-step solution:
### Step 1: Understand the equations
The hyperbola is given by:
\[
xy = 4
\]
The line can be rewritten as:
\[
y = -2x + 4
\]
### Step 2: Find the parametric form of the hyperbola
For the hyperbola \(xy = 4\), we can express points on the hyperbola in parametric form. Let:
\[
x = 2t \quad \text{and} \quad y = \frac{4}{2t} = \frac{2}{t}
\]
Thus, a point \(P\) on the hyperbola can be represented as:
\[
P(2t, \frac{2}{t})
\]
### Step 3: Find the equation of the tangent line at point P
The equation of the tangent line to the hyperbola at point \(P(2t, \frac{2}{t})\) can be derived using the formula:
\[
x + y \cdot t^2 = 4
\]
Substituting \(P\) into the tangent equation gives:
\[
x + \frac{2}{t} \cdot t^2 = 4 \implies x + 2t = 4
\]
This simplifies to:
\[
x + 2t = 4 \implies y = -2t + 4
\]
### Step 4: Find the intercepts of the tangent line
To find the x-intercept (A), set \(y = 0\):
\[
0 = -2t + 4 \implies 2t = 4 \implies t = 2
\]
Thus, the x-intercept is:
\[
A(4, 0)
\]
To find the y-intercept (B), set \(x = 0\):
\[
y = -2t + 4 \implies y = 4 - 2t
\]
So the y-intercept is:
\[
B(0, 4 - 2t)
\]
### Step 5: Find the slope of the tangent line
The slope of the tangent line can be found from the equation:
\[
\frac{dy}{dx} = -2
\]
### Step 6: Find the slope of the line
The slope of the line \(y + 2x = 4\) is:
\[
\frac{dy}{dx} = -2
\]
### Step 7: Find the common normal
Since the slopes of the tangent and the line are equal, we can find the point where the common normal intersects. The slopes are equal, so we set:
\[
-t^2 = -\frac{1}{2} \implies t^2 = \frac{1}{2} \implies t = \frac{1}{\sqrt{2}}
\]
### Step 8: Find the coordinates of point P
Substituting \(t = \frac{1}{\sqrt{2}}\) into the parametric equations gives:
\[
P\left(2 \cdot \frac{1}{\sqrt{2}}, \frac{2}{\frac{1}{\sqrt{2}}}\right) = P(\sqrt{2}, 2\sqrt{2})
\]
### Step 9: Calculate the shortest distance
The shortest distance \(D\) from point \(P\) to the line can be calculated using the formula for the distance from a point to a line:
\[
D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
For the line \(y + 2x - 4 = 0\), \(A = 2\), \(B = 1\), and \(C = -4\). The coordinates of \(P\) are \((\sqrt{2}, 2\sqrt{2})\):
\[
D = \frac{|2(\sqrt{2}) + 1(2\sqrt{2}) - 4|}{\sqrt{2^2 + 1^2}} = \frac{|2\sqrt{2} + 2\sqrt{2} - 4|}{\sqrt{5}} = \frac{|4\sqrt{2} - 4|}{\sqrt{5}} = \frac{4(\sqrt{2} - 1)}{\sqrt{5}}
\]
### Final Answer
Thus, the shortest distance between the hyperbola and the line is:
\[
D = \frac{4(\sqrt{2} - 1)}{\sqrt{5}}
\]
