The diagram shows part of the graph of $y = 2\sqrt{2 + x}$. It cuts the x-axis at $(-2,0)$. The line $y = 2x$ meets the curve at $P$. The normal to the curve at $P$ meets the x-axis at $R$.

1. Find the coordinates of $P$ by equating the line and the curve

$2x = 2\sqrt{2 + x}$
$x = \sqrt{2 + x}$
$x^2 = 2 + x$
$x^2 - x - 2 = 0$
$(x - 2)(x + 1) = 0$
Since $x \ge 0$ for $x = \sqrt{2+x}$, $x = 2$.
When $x = 2$, $y = 2(2) = 4$.
So, $P$ is $(2, 4)$.

2. Find the gradient function of the curve

$y = 2(2 + x)^{\frac{1}{2}}$
$\frac{dy}{dx} = 2 \left( \frac{1}{2} \right) (2 + x)^{-\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{\sqrt{2 + x}}$

3. Calculate the gradient of the tangent and normal at $P$

At $x = 2$, gradient of tangent $m_t = \frac{1}{\sqrt{2 + 2}} = \frac{1}{2}$.
Gradient of normal $m_n = -\frac{1}{m_t} = -2$.

4. Find the equation of the normal at $P$

$y - 4 = -2(x - 2)$
$y = -2x + 4 + 4$
$y = -2x + 8$

5. Find the coordinates of $R$ by setting $y = 0$

At $R$, the normal meets the x-axis, so $y = 0$.
$0 = -2x + 8$
$2x = 8$
$x = 4$
The coordinates of $R$ are $(4, 0)$.

1. Find the coordinates of $P$ by solving the equations of the curve and the line simultaneously.

$2x = 2\sqrt{2 + x}$
$x = \sqrt{2 + x}$
$x^2 = 2 + x$
$x^2 - x - 2 = 0$
$(x - 2)(x + 1) = 0$
Since $P$ is in the first quadrant, $x = 2$.
$y = 2(2) = 4$
So, $P$ is $(2, 4)$.

2. Find the coordinates of $R$ by finding the equation of the normal at $P$.

$y = 2(2 + x)^{\frac{1}{2}}$
$\frac{dy}{dx} = 2 \left(\frac{1}{2}\right)(2 + x)^{-\frac{1}{2}} = \frac{1}{\sqrt{2 + x}}$
At $x = 2$, gradient of tangent $m_t = \frac{1}{\sqrt{4}} = \frac{1}{2}$.
Gradient of normal $m_n = -\frac{1}{m_t} = -2$.
Equation of normal at $P(2, 4)$:
$y - 4 = -2(x - 2)$
$y = -2x + 8$
At $R$, $y = 0$:
$0 = -2x + 8 \implies x = 4$.
So, $R$ is $(4, 0)$.

3. Calculate the lengths of the sides of triangle $OPR$ to show it is isosceles.

Coordinates are $O(0, 0)$, $P(2, 4)$, and $R(4, 0)$.
Length of $OP = \sqrt{(2 - 0)^2 + (4 - 0)^2} = \sqrt{4 + 16} = \sqrt{20}$
Length of $PR = \sqrt{(4 - 2)^2 + (0 - 4)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20}$
Length of $OR = 4$
Since $OP = PR = \sqrt{20}$, triangle $OPR$ has two equal sides and is therefore an isosceles triangle.