To solve the problem, we need to find the area of triangle QPF formed by the points Q, P, and F, where P is the point on the curve, R is the point where the tangent at P intersects the x-axis, and Q is the point on the latus rectum of the hyperbola defined by the equation \(x^2 - 2y^2 = 4\).
### Step 1: Identify the curve and its properties
The given curve is \(x^2 - 2y^2 = 4\). This is a hyperbola. We can rewrite it in standard form:
\[
\frac{x^2}{4} - \frac{y^2}{2} = 1
\]
From this, we can identify:
- The semi-major axis \(a = 2\)
- The semi-minor axis \(b = \sqrt{2}\)
### Step 2: Find the focus of the hyperbola
The distance of the foci from the center is given by \(c = \sqrt{a^2 + b^2}\):
\[
c = \sqrt{4 + 2} = \sqrt{6}
\]
Thus, the foci are located at \((\pm \sqrt{6}, 0)\).
### Step 3: Determine the coordinates of point P
Point \(P\) is given as \((4, \sqrt{6})\).
### Step 4: Find the equation of the tangent at point P
The equation of the tangent to the hyperbola at point \((x_0, y_0)\) is given by:
\[
\frac{xx_0}{4} - \frac{yy_0}{2} = 1
\]
Substituting \(P(4, \sqrt{6})\):
\[
\frac{4x}{4} - \frac{\sqrt{6}y}{2} = 1 \implies x - \frac{\sqrt{6}}{2}y = 1
\]
### Step 5: Find the intersection point R with the x-axis
To find the intersection with the x-axis, set \(y = 0\):
\[
x - 0 = 1 \implies x = 1
\]
Thus, point \(R\) is \((1, 0)\).
### Step 6: Find the coordinates of point Q on the latus rectum
The latus rectum of the hyperbola is given by the lines \(x = \pm 2\) (since \(2a = 4\)). We will consider the right latus rectum at \(x = 2\):
Substituting \(x = 2\) into the hyperbola equation:
\[
2^2 - 2y^2 = 4 \implies 4 - 2y^2 = 4 \implies -2y^2 = 0 \implies y = 0
\]
Thus, point \(Q\) is \((2, 0)\).
### Step 7: Identify the focus F nearest to P
The nearest focus to point \(P(4, \sqrt{6})\) is \(F(\sqrt{6}, 0)\).
### Step 8: Calculate the area of triangle QPF
The area \(A\) of triangle formed by points \(Q(2, 0)\), \(P(4, \sqrt{6})\), and \(F(\sqrt{6}, 0)\) can be calculated using the formula:
\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
A = \frac{1}{2} \left| 2(\sqrt{6} - 0) + 4(0 - 0) + \sqrt{6}(0 - \sqrt{6}) \right|
\]
\[
= \frac{1}{2} \left| 2\sqrt{6} - 6 \right| = \frac{1}{2} \left| 2\sqrt{6} - 6 \right|
\]
### Step 9: Simplify the area expression
Calculating the area:
\[
A = \frac{1}{2} \left| 2\sqrt{6} - 6 \right| = \frac{1}{2} (6 - 2\sqrt{6}) \text{ (since } 2\sqrt{6} < 6 \text{)}
\]
Thus, the area of triangle \(QPF\) is:
\[
A = 3 - \sqrt{6}
\]
### Final Answer
The area of triangle \(QPF\) is \(3 - \sqrt{6}\).
