Common Tangent Interval Calculation

This solution determines the possible values of k for which the curves $y^2 = kx$ and $xy = -1$ share a common tangent line with a slope of $\frac{1}{2}$, and identifies the interval where k cannot lie.

Parabola Tangent Analysis ($y^2 = kx$)

Differentiate $y^2 = kx$ implicitly with respect to $x$:

$2y \frac{dy}{dx} = k \implies \frac{dy}{dx} = \frac{k}{2y}$

Given the slope $m = \frac{1}{2}$, we set:

$\frac{k}{2y} = \frac{1}{2} \implies y = k$

Substitute $y=k$ back into the parabola's equation ($y^2=kx$):

$k^2 = kx$

Since $k=0$ is inconsistent with the hyperbola ($xy=-1$), we know $k \neq 0$. Thus, $x=k$.

The point of tangency on the parabola is $(k, k)$.

The equation of the tangent line is $y = mx + c$. Using the point $(k,k)$ and slope $m = \frac{1}{2}$:

$k = \frac{1}{2}(k) + c \implies c = k - \frac{k}{2} = \frac{k}{2}$

The tangent line to the parabola is $y = \frac{1}{2}x + \frac{k}{2}$.

Hyperbola Tangent Analysis ($xy = -1$)

Differentiate $xy = -1$ implicitly with respect to $x$:

$y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}$

Set the slope $\frac{dy}{dx} = \frac{1}{2}$:

$-\frac{y}{x} = \frac{1}{2} \implies y = -\frac{x}{2}$

Substitute $y = -\frac{x}{2}$ back into the hyperbola's equation ($xy=-1$):

$x\left(-\frac{x}{2}\right) = -1 \implies -\frac{x^2}{2} = -1 \implies x^2 = 2 \implies x = \pm \sqrt{2}$

Find the corresponding $y$ values and points of tangency:

• If $x = \sqrt{2}$, then $y = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$. Point: $(\sqrt{2}, -1/\sqrt{2})$.
• If $x = -\sqrt{2}$, then $y = -\frac{-\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Point: $(-\sqrt{2}, 1/\sqrt{2})$.

The tangent line to the hyperbola $xy = -1$ at a point $(x_0, y_0)$ is $y x_0 + x y_0 = -2$. Rewriting in slope-intercept form ($y=mx+c$):

$y = -\frac{y_0}{x_0}x - \frac{2}{x_0}$

The slope $m = -\frac{y_0}{x_0}$ is given as $\frac{1}{2}$. The y-intercepts ($c$) for the two possible tangent points are:

• For $(\sqrt{2}, -1/\sqrt{2})$: $c = - \frac{-1/\sqrt{2}}{\sqrt{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}$. The tangent line is $y = \frac{1}{2}x - \sqrt{2}$.
• For $(-\sqrt{2}, 1/\sqrt{2})$: $c = - \frac{1/\sqrt{2}}{-\sqrt{2}} = -\frac{2}{-\sqrt{2}} = \sqrt{2}$. The tangent line is $y = \frac{1}{2}x + \sqrt{2}$.

Equating Tangent Conditions

For a common tangent, the y-intercepts derived from both curve analyses must be equal. The tangent to the parabola has a y-intercept $c = \frac{k}{2}$, while the tangents to the hyperbola have y-intercepts $c = \pm \sqrt{2}$.

Equating these y-intercepts:

$\frac{k}{2} = \sqrt{2} \quad \text{or} \quad \frac{k}{2} = -\sqrt{2}$

Solving for $k$, the possible values are:

$k = 2\sqrt{2} \quad \text{or} \quad k = -2\sqrt{2}$

Numerically, these values are approximately $k \approx 2.828$ and $k \approx -2.828$.

Identifying k's Non-Interval

The question asks for the interval where k *cannot* lie. The possible values for $k$ are $2\sqrt{2}$ and $-2\sqrt{2}$. We examine the given options:

• Option 1: $(2, 4]$ contains $k \approx 2.828$.
• Option 2: $(-3, -1)$ contains $k \approx -2.828$.
• Option 4: $(1, 3)$ contains $k \approx 2.828$.
• Option 3: $[-4, -3)$. This interval includes values of $k$ such that $-4 \le k < -3$.
  • $k \approx 2.828$ is not in $[-4, -3)$.
  • $k \approx -2.828$ is also not in $[-4, -3)$, because $-2.828$ is greater than $-3$.

Since neither of the possible values for $k$ ($2\sqrt{2}$ or $-2\sqrt{2}$) falls within the interval $[-4, -3)$, this is the interval where k cannot lie.