写出将方程为 $y^{2}-x^{2}=1$ 的曲线变换为方程为 $9y^{2}-54y - x^{2} - 2x + 79 = 0.$ 的曲线的一系列变换步骤。 曲线 $C$ 的方程为 $9y^{2}-54y - x^{2} - 2x + 79 = 0.$ 对于实数 $x$，使用非几何作图的方法确定 $y$ 不能介于 $a$ 和 $b$ 之间，其中 $a$ 和 $b$ 是待求的精确实常数。 画出曲线 $C$ 的草图，清晰地标出所有渐近线的方程以及极值点的坐标。 通过添加一条合适的曲线，确定方程 $9\bigl[(x+1)^{2}+3\bigr]^{2} - 54\bigl[(x+1)^{2}+3\bigr] - x^{2} - 2x + 79 = 0.$ 的实根个数。

1. Complete the square for both $x$ and $y$ in the given equation to rewrite it in standard form.

$9y^{2}-54y - x^{2} - 2x + 79 = 0$
$9(y^{2}-6y) - (x^{2}+2x) + 79 = 0$
$9\bigl[(y-3)^{2}-9\bigr] - \bigl[(x+1)^{2}-1\bigr] + 79 = 0$
$9(y-3)^{2} - 81 - (x+1)^{2} + 1 + 79 = 0$
$9(y-3)^{2} - (x+1)^{2} = 1$

2. Rewrite the terms to explicitly reveal the transformations applied to the original equation $y^2 - x^2 = 1$.

We can rewrite the transformed equation as:
$[3(y-3)]^{2} - (x+1)^{2} = 1$
Comparing this with the original equation $y^2 - x^2 = 1$, we can obtain the new equation by applying changes to $x$ and $y$ sequentially:
First, replace $y$ with $3y$, giving $(3y)^2 - x^2 = 1$.
Then, replace $x$ with $x+1$ and $y$ with $y-3$, giving $[3(y-3)]^2 - (x+1)^2 = 1$.

3. Translate the algebraic replacements into a geometric sequence of transformations.

Replacing $y$ with $3y$ corresponds to a scaling parallel to the $y$-axis by a scale factor of $\frac{1}{3}$.
Replacing $x$ with $x+1$ and $y$ with $y-3$ corresponds to a translation of $1$ unit in the negative $x$-direction and $3$ units in the positive $y$-direction.

Hence, the sequence is:

1. Scaling parallel to the $y$-axis by a scale factor of $\frac{1}{3}$.
2. Translation by vector $\begin{pmatrix} -1 \\ 3 \end{pmatrix}$.

1. Express the curve equation in completed square form and isolate the $x$ term.

By completing the square (as is naturally done to identify the transformations in part (a)):
$9(y^{2}-6y) - (x^{2}+2x) + 79 = 0$
$9[(y-3)^{2}-9] - [(x+1)^{2}-1] + 79 = 0$
$9(y-3)^{2} - 81 - (x+1)^{2} + 1 + 79 = 0$
$9(y-3)^{2} - (x+1)^{2} = 1$
Rearranging to isolate the squared $x$ term:
$(x+1)^{2} = 9(y-3)^{2} - 1$

2. Apply the non-negativity property of squares for real values of $x$.

For $x$ to be a real number, its square must be non-negative:
$(x+1)^{2} \ge 0$
Therefore, substitute the expression for $(x+1)^{2}$:
$9(y-3)^{2} - 1 \ge 0$
$9(y-3)^{2} \ge 1$
$(y-3)^{2} \ge \frac{1}{9}$

3. Solve the inequality to determine the restricted range for $y$ and state the constants $a$ and $b$.

Taking the square root of both sides gives a modulus inequality:
$|y-3| \ge \frac{1}{3}$
$y-3 \le -\frac{1}{3} \quad \text{or} \quad y-3 \ge \frac{1}{3}$
$y \le \frac{8}{3} \quad \text{or} \quad y \ge \frac{10}{3}$
This means $y$ cannot lie strictly between $\frac{8}{3}$ and $\frac{10}{3}$.
Hence, $a = \frac{8}{3}$ and $b = \frac{10}{3}$.

1. Express the equation in standard form to find the center and turning points.

Completing the square (or using the form from part (a)):
$9(y^2 - 6y) - (x^2 + 2x) + 79 = 0$
$9(y - 3)^2 - 81 - (x + 1)^2 + 1 + 79 = 0$
$9(y - 3)^2 - (x + 1)^2 = 1$
This is a hyperbola opening upwards and downwards. Its center is at $(-1, 3)$.
The turning points occur when $x = -1$:
$9(y - 3)^2 = 1 \implies y - 3 = \pm \frac{1}{3}$
$y = \frac{10}{3} \text{ or } y = \frac{8}{3}$
The turning points are $\left(-1, \frac{10}{3}\right)$ and $\left(-1, \frac{8}{3}\right)$.

2. Find the equations of the asymptotes.

The asymptotes are found by setting the constant on the RHS to 0:
$9(y - 3)^2 - (x + 1)^2 = 0$
$3(y - 3) = \pm(x + 1)$
Solving for $y$, we get the two asymptotes:
$3y - 9 = x + 1 \implies y = \frac{1}{3}x + \frac{10}{3}$
$3y - 9 = -(x + 1) \implies y = -\frac{1}{3}x + \frac{8}{3}$

3. Sketch the curve.

Draw a set of coordinate axes.
Sketch the asymptotes $y = \frac{1}{3}x + \frac{10}{3}$ and $y = -\frac{1}{3}x + \frac{8}{3}$ as dotted lines intersecting at $(-1, 3)$.
Draw the upper branch of the hyperbola above $y = \frac{10}{3}$ with its minimum at $\left(-1, \frac{10}{3}\right)$, approaching the asymptotes as $x \to \pm\infty$.
Draw the lower branch of the hyperbola below $y = \frac{8}{3}$ with its maximum at $\left(-1, \frac{8}{3}\right)$, approaching the asymptotes as $x \to \pm\infty$.

1. Identify the equation of the suitable curve to add.

Comparing the given equation $9\bigl[(x+1)^{2}+3\bigr]^{2} - 54\bigl[(x+1)^{2}+3\bigr] - x^{2} - 2x + 79 = 0$ with the equation of curve $C$, $9y^{2}-54y - x^{2} - 2x + 79 = 0$, we observe that $y$ has been replaced by $(x+1)^2 + 3$.
Thus, the suitable curve to add to the sketch is the parabola $y = (x+1)^2 + 3$.

2. Determine the number of intersections by considering the graphical features of both curves.

From part (b)(ii), the hyperbola $C$ has an upper branch with a minimum vertex at $(-1, \frac{10}{3})$ and a lower branch with a maximum vertex at $(-1, \frac{8}{3})$.
The added curve $y = (x+1)^2 + 3$ is an upward-opening parabola with its minimum point at $(-1, 3)$.
Since the minimum $y=3$ is strictly between $\frac{8}{3}$ and $\frac{10}{3}$, the parabola's vertex lies below the upper branch of $C$ and above the lower branch.
Because the parabola opens upwards and its arms grow quadratically (faster than the hyperbola, which approaches linear asymptotes), the parabola will cross the upper branch of $C$ on both sides of the axis of symmetry $x = -1$.
Since the parabola lies entirely above $y=3$, it will never intersect the lower branch.

3. State the final number of real roots.

Since there are exactly 2 points of intersection between the two graphs, the equation has 2 real roots.