上图显示了当 $x>0$ 时，方程为 $e^{2y} = x^2 + y$ 的曲线 $C$。

1. Determine the symmetry of the curve from its equation

Substitute $-x$ for $x$ in the curve's equation:
$e^{2y} = (-x)^2 + y$
$e^{2y} = x^2 + y$
Since the equation remains unchanged when $x$ is replaced by $-x$, the curve $C$ is symmetrical about the $y$-axis.

2. Draw the required portion of the curve

Reflect the provided portion of $C$ (where $x > 0$) across the $y$-axis. Since the original curve passes through $(1, 0)$, the reflected portion must pass through $(-1, 0)$.

1. Differentiate both sides of the equation with respect to $x$.

$\frac{d}{dx}(e^{2y}) = \frac{d}{dx}(x^2 + y)$
$2e^{2y} \frac{dy}{dx} = 2x + \frac{dy}{dx}$

2. Make $\frac{dy}{dx}$ the subject of the formula.

$2e^{2y} \frac{dy}{dx} - \frac{dy}{dx} = 2x$
$(2e^{2y} - 1) \frac{dy}{dx} = 2x$
$\frac{dy}{dx} = \frac{2x}{2e^{2y}-1}$

1. Identify the condition for a vertical tangent.

The line $x = k$ is a vertical line. For this line to be a tangent to the curve, the gradient $\frac{dy}{dx}$ must be undefined. Therefore, the denominator of $\frac{dy}{dx}$ must be zero:
$2e^{2y} - 1 = 0$

2. Solve for the $y$-coordinate of the point of tangency.

$e^{2y} = \frac{1}{2}$
Taking the natural logarithm on both sides:
$2y = \ln\left(\frac{1}{2}\right) = -\ln 2$
$y = -\frac{1}{2}\ln 2$

3. Substitute $y$ and $e^{2y}$ back into the original curve equation to find $x^2$.

Substitute $e^{2y} = \frac{1}{2}$ and $y = -\frac{1}{2}\ln 2$ into the equation of $C$, $e^{2y} = x^2 + y$:
$\frac{1}{2} = x^2 + \left(-\frac{1}{2}\ln 2\right)$
$x^2 = \frac{1}{2} + \frac{1}{2}\ln 2$
$x^2 = \frac{1 + \ln 2}{2}$

4. Solve for $x$ to determine the values of $k$.

Since the vertical tangent corresponds to the line $x = k$, we have $x = k$ at this point of tangency:
$k^2 = \frac{1 + \ln 2}{2}$
$k = \pm \sqrt{\frac{1 + \ln 2}{2}}$

1. Find the gradient of the tangent at $(1, 0)$

Substitute $x = 1$ and $y = 0$ into the derivative:
$\frac{dy}{dx} = \frac{2(1)}{2e^{2(0)}-1} = \frac{2}{2(1)-1} = 2$

2. Find the equation of the tangent

Using the point-gradient form with point $(1, 0)$ and gradient $m = 2$:
$y - 0 = 2(x - 1)$
$y = 2x - 2$

3. Set up the equation to find where the tangent meets the curve again

Substitute the tangent equation $y = 2x - 2$ into the equation of curve $C$:
$e^{2(2x-2)} = x^2 + (2x - 2)$
$e^{4x-4} = x^2 + 2x - 2$

4. Solve for $x$ using a Graphing Calculator

Plot $y = e^{4x-4}$ and $y = x^2 + 2x - 2$ on a graphing calculator and locate their points of intersection.
The solutions are $x = 1$ (the known point of tangency) and $x \approx -2.73205$.
The $x$-coordinate where the tangent meets $C$ again is $-2.73$ (to 3 s.f.).