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Step-by-Step Solution
Step 1: Identify the given curves and important points
We have the parabola
$P: y^2 = 4x$
whose vertex is at
$O(0,0)$
and focus at
$F_P(1,0)$.
We also have the hyperbola
$H: x^2 - y^2 = 4$
which can be rewritten as
$ \frac{x^2}{4} - \frac{y^2}{4} = 1.$
Its semi-major axis is
$a = 2,$
and the coordinates of its focus on the positive $x$-axis are
$F_H(2\sqrt{2}, 0).$
Step 2: Determine the equation of the line L
The line
$L: y = mx + c,$
with slope
$m > 0,$
is given to be:
A focal chord of the parabola $y^2 = 4x$, so it must pass through the focus of the parabola $F_P(1,0)$, hence
$0 = m \cdot 1 + c \implies c = -m.$
A tangent to the hyperbola $x^2 - y^2 = 4.$
Thus, the line can be written as
$y = m x - m.$
Step 3: Impose the tangency condition to the hyperbola
For the hyperbola $x^2 - y^2 = 4,$ the condition for the line
$y = mx + c$
to be tangent is obtained by substituting
$y = mx + c$
into the hyperbola and ensuring the resulting quadratic has a single (repeated) root. Let us outline the standard procedure:
1. Write the hyperbola in the form
$\frac{x^2}{4} - \frac{y^2}{4} = 1.$
2. Substitute $y = mx + c$ into
$x^2 - y^2 = 4:$
\[
x^2 - (mx + c)^2 = 4.
\]
3. The condition for tangency is that the quadratic in $x$ has a single solution. Equivalently, its discriminant must be zero.
In the solution provided, it is deduced that
$c = \pm \sqrt{4 m^2 - 4}.$
Since we already have $c = -m,$ we set
$-m = \pm \sqrt{4 m^2 - 4}.$
Squaring both sides gives
$m^2 = 4m^2 - 4 \implies 3m^2 = 4 \implies m^2 = \frac{4}{3} \implies m = \frac{2}{\sqrt{3}}$
(taking the positive root since $m > 0$).
Hence
$c = -m = -\frac{2}{\sqrt{3}}.$
Therefore, the line $L$ is
\[
y = \frac{2}{\sqrt{3}}\,x \;-\; \frac{2}{\sqrt{3}}.
\]
Step 4: Find the intersection of L with the parabola P
The parabola is
$y^2 = 4x.$
From this,
$x = \frac{y^2}{4}.$
Substitute $x = \frac{y^2}{4}$ in the line equation
$y = \frac{2}{\sqrt{3}} x \;-\;\frac{2}{\sqrt{3}}.$
Substituting
$x = \frac{y^2}{4}$
into
$y = \frac{2}{\sqrt{3}} \left(\frac{y^2}{4}\right) - \frac{2}{\sqrt{3}},$
we get:
\[
y
= \frac{2}{\sqrt{3}} \cdot \frac{y^2}{4}
- \frac{2}{\sqrt{3}}
= \frac{y^2}{2\sqrt{3}}
- \frac{2}{\sqrt{3}}.
\]
Rearranging, multiply through by $2\sqrt{3}$ to clear denominators:
\[
2\sqrt{3} \, y
= y^2
- 4.
\]
Rewrite to standard quadratic form:
\[
y^2 \;-\; 2\sqrt{3}\,y \;-\; 4 = 0.
\]
Let the roots be $y_1$ and $y_2.$ These correspond to the $y$-coordinates of the intersection points $M$ and $N$ on the parabola.
Step 5: Coordinates of M and N
If $y_1$ and $y_2$ are roots of
$y^2 - 2\sqrt{3}\,y - 4 = 0,$
then their corresponding $x$-coordinates on the parabola are
$x_1 = \frac{y_1^2}{4}\,$ and $\,x_2 = \frac{y_2^2}{4}.$
Hence the intersection points are
\[
M(x_1,\,y_1),
\quad
N(x_2,\,y_2).
\]
Step 6: Identify the points O and F in the quadrilateral OMFN
The quadrilateral OMFN has vertices:
$O(0,0)$ : \,vertex of the parabola.
$M(x_1,y_1)$ : \,intersection of line $L$ with the parabola.
$F(2\sqrt{2}, 0)$ : \,focus of the hyperbola $x^2 - y^2 = 4$ on the positive $x$-axis.
$N(x_2,y_2)$ : \,other intersection of line $L$ with the parabola.
Step 7: Formula for the area of quadrilateral OMFN
One way to find the area of the quadrilateral
$O(0,0) - M(x_1,y_1) - F(2\sqrt{2},0) - N(x_2,y_2)$
is to use the shoelace formula or to split it into triangles
$OMF$
and
$OFN.$
Using a determinant-based area approach for polygon vertices in order
$(0,0)\rightarrow M(x_1,y_1)\rightarrow F(2\sqrt{2},0)\rightarrow N(x_2,y_2)\rightarrow (0,0),$
the expression for area simplifies to something proportional to
$\left|\,y_2 - y_1\right|.$
Step 8: Simplify using the quadratic's properties
From the given (and verified) computations, the area turns out to be:
\[
\text{Area}
= \sqrt{2}\,\bigl|\,y_2 - y_1\bigr|.
\]
Using the relationship between the roots $y_1$ and $y_2$ of the quadratic
$y^2 - 2\sqrt{3}\,y - 4 = 0,$
we use:
\[
y_1 + y_2 = 2\sqrt{3},
\quad
y_1\,y_2 = -4.
\]
Then
\[
(y_2 - y_1)^2
= (y_1 + y_2)^2 - 4\,y_1y_2
= (2\sqrt{3})^2 - 4(-4)
= 12 + 16
= 28.
\]
Hence
$|y_2 - y_1| = \sqrt{28} = 2\sqrt{7}.$
Thus,
\[
\text{Area}
= \sqrt{2} \,\times 2\sqrt{7}
= 2\sqrt{14}.
\]
Step 9: Conclusion
Therefore, the area of the quadrilateral $OMFN$ is
$\boxed{2\sqrt{14}}.$
