The curve $C$ has equation
$y = \frac{x^2 + 4x + \lambda - 5}{x + r},$
where $\lambda$ is a non-zero constant, and a vertical asymptote $x = 1$.

1. State the value of $r$ using the vertical asymptote

The vertical asymptote occurs when the denominator is zero.
Given the vertical asymptote is $x = 1$, we have $x + r = 0$ when $x = 1$.
$1 + r = 0 \implies r = -1$

2. Rewrite the equation of $C$ by polynomial division to find the other asymptote

Substitute $r = -1$ into the equation of $C$:
$y = \frac{x^2 + 4x + \lambda - 5}{x - 1}$
By polynomial long division or synthetic division, we divide the numerator by $x - 1$:
$x^2 + 4x + \lambda - 5 = x(x - 1) + 5x + \lambda - 5$
$= x(x - 1) + 5(x - 1) + 5 + \lambda - 5$
$= (x + 5)(x - 1) + \lambda$
Thus, the equation can be written as:
$y = x + 5 + \frac{\lambda}{x - 1}$

3. State the equation of the oblique asymptote

As $x \to \pm\infty$, the fractional term $\frac{\lambda}{x - 1} \to 0$.
Therefore, the equation of the other asymptote (oblique asymptote) is:
$y = x + 5$

1. Recall the exact form of the curve and its asymptotes from part (i).

Using $r = -1$ and performing algebraic division as in part (i), we have:
$y = \frac{x^2 + 4x + \lambda - 5}{x - 1} = x + 5 + \frac{\lambda}{x - 1}$
The asymptotes are the vertical line $x = 1$ and the oblique line $y = x + 5$.

2. Determine the shape and position of the curve relative to the asymptotes for $\lambda < 0$.

Since $\lambda < 0$:
For $x < 1$, $x - 1 < 0 \implies \frac{\lambda}{x - 1} > 0$. The curve lies above the oblique asymptote.
For $x > 1$, $x - 1 > 0 \implies \frac{\lambda}{x - 1} < 0$. The curve lies below the oblique asymptote.

Differentiating gives $y' = 1 - \frac{\lambda}{(x-1)^2}$. Since $\lambda < 0$, $-\frac{\lambda}{(x-1)^2} > 0$, so $y' > 0$ for all $x \neq 1$. The curve is strictly increasing with no stationary points.

3. Find the coordinates of the axial intercepts.

For the $y$-intercept, set $x = 0$:
$y = \frac{0 + 0 + \lambda - 5}{0 - 1} = 5 - \lambda$
Since $\lambda < 0$, the $y$-intercept is positive and greater than 5.

For the $x$-intercepts, set $y = 0$:
$x^2 + 4x + \lambda - 5 = 0$
Using the quadratic formula:
$x = \frac{-4 \pm \sqrt{16 - 4(1)(\lambda - 5)}}{2} = \frac{-4 \pm \sqrt{36 - 4\lambda}}{2} = -2 \pm \sqrt{9 - \lambda}$
(One intercept is positive and one is negative).

1. Set up the intersection equation using $r=-1$ from part (i).

From part (i), since $x=1$ is a vertical asymptote, the denominator is zero at $x=1$, so $r = -1$. Equating the line and the curve for intersections:
$\frac{x^2 + 4x + \lambda - 5}{x - 1} = 2x$

2. Rearrange the equation into standard quadratic form.

Multiply both sides by $(x - 1)$ (assuming $x \neq 1$):
$x^2 + 4x + \lambda - 5 = 2x(x - 1)$
$x^2 + 4x + \lambda - 5 = 2x^2 - 2x$
$x^2 - 6x + 5 - \lambda = 0$

3. Use the discriminant to find the condition for at least one point in common.

For the line and curve to have at least one point of intersection, the quadratic equation must have real roots. Thus, its discriminant must be non-negative ($\Delta \ge 0$):
$b^2 - 4ac \ge 0$
$(-6)^2 - 4(1)(5 - \lambda) \ge 0$
$36 - 20 + 4\lambda \ge 0$
$16 + 4\lambda \ge 0$
$4\lambda \ge -16$
$\lambda \ge -4$

4. Account for the given restriction that $\lambda$ is non-zero to state the final range.

We must verify if an intersection could ever occur at the vertical asymptote $x=1$. Substituting $x=1$ into the quadratic equation gives:
$(1)^2 - 6(1) + 5 - \lambda = 0 \implies \lambda = 0$
Since we are given in the question stem that $\lambda$ is a non-zero constant, $x=1$ will never be a root, and $\lambda=0$ must be excluded from the valid range.
Therefore, the required range of values is:
$\lambda \ge -4, \lambda \neq 0$

1. Substitute the given values to form the equation of the curve and state its asymptotes.

From part (i), we know $r = -1$. Substituting $\lambda = 9$, the equation of the curve $C$ becomes:
$y = \frac{x^2 + 4x + 4}{x - 1} = \frac{(x+2)^2}{x-1}$
By polynomial division (or from part (i)), this can be written as:
$y = x + 5 + \frac{9}{x - 1}$
The asymptotes are the vertical line $x = 1$ and the oblique line $y = x + 5$.

2. Determine the axis intercepts.

For the $y$-intercept, set $x = 0$:
$y = \frac{0^2 + 4(0) + 4}{0 - 1} = -4$
The $y$-intercept is $(0, -4)$.

For the $x$-intercept, set $y = 0$:
$\frac{(x+2)^2}{x-1} = 0 \implies (x+2)^2 = 0 \implies x = -2$
The $x$-intercept is $(-2, 0)$.

3. Find the coordinates of the stationary points.

Differentiating $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( x + 5 + \frac{9}{x - 1} \right) = 1 - \frac{9}{(x-1)^2}$
At stationary points, $\frac{dy}{dx} = 0$:
$1 - \frac{9}{(x-1)^2} = 0$
$(x-1)^2 = 9$
$x - 1 = 3 \quad \text{or} \quad x - 1 = -3$
$x = 4 \quad \text{or} \quad x = -2$
When $x = 4$, $y = \frac{(4+2)^2}{4-1} = \frac{36}{3} = 12$.
When $x = -2$, $y = \frac{(-2+2)^2}{-2-1} = 0$.
The stationary points are a local minimum at $(4, 12)$ and a local maximum at $(-2, 0)$.

4. Sketch the graph incorporating all determined features.

Draw the required sketch on a Cartesian plane, ensuring:

• The vertical asymptote $x = 1$ and oblique asymptote $y = x + 5$ are drawn as dashed lines and labelled.
• The left branch of the curve passes through the $y$-intercept $(0, -4)$ and has a local maximum at $(-2, 0)$ (touching the $x$-axis).
• The right branch of the curve is situated above the oblique asymptote and has a local minimum at $(4, 12)$.
• The appropriate asymptotic behaviour is shown at the tails of both branches.

1. Identify the shape and center of curve $D$ when $k=3$.

Substitute $k = 3$ into the given equation of $D$:
$\frac{(x-4)^2}{4^2} + \frac{(y-6)^2}{3^2} = 1$
This is the standard equation of an ellipse with its center at $(4, 6)$.

2. Determine the vertices and relative position to sketch $D$.

The horizontal semi-axis is $a = 4$ and the vertical semi-axis is $b = 3$.
The vertices are $(4 \pm 4, 6)$ and $(4, 6 \pm 3)$, yielding the points $(0, 6)$, $(8, 6)$, $(4, 9)$, and $(4, 3)$.
On the sketch, the ellipse is tangent to the $y$-axis at $(0, 6)$. Its highest point, $(4, 9)$, is vertically aligned with but completely below the local minimum of curve $C$ at $(4, 12)$.

1. Identify the relevant stationary point of $C$ and the highest point of the ellipse $D$.

From part (iv), the right branch of the curve $C$ (where $x > 1$) has a U-shape with a minimum point at $(4, 12)$.
The curve $D$ is an ellipse with its centre at $(4, 6)$. The maximum $y$-coordinate on $D$ occurs at the same $x$-coordinate ($x = 4$), which yields $y = 6 + |k|$.

2. Set up an inequality for $C$ and $D$ to intersect more than once and solve for $k$.

At $x = 4$, the curve $C$ is U-shaped (concave upwards) and the top of the ellipse $D$ is an inverted U-shape (concave downwards).
If $D$ touches $C$ exactly at $(4, 12)$, they intersect exactly once. For $C$ and $D$ to intersect more than once (i.e. to cross each other to form at least two intersection points), the peak of $D$ must strictly exceed the minimum point of $C$:
$6 + |k| > 12$
$|k| > 6$
$k < -6 \text{ or } k > 6$