For which values of the parameter lambda can tangent

For which values of the parameter $\lambda$ can tangent lines to the curve $y=(x+\lambda)e^{x}$ be drawn from the origin $(0,0)$?

Step 1: The tangency condition

Let $T=(t,(t+\lambda)e^{t})$ be an arbitrary point on the curve. Since $y'=(x+\lambda+1)e^{x}$, the slope at $T$ is $k=(t+\lambda+1)e^{t}$, and the tangent is $y-(t+\lambda)e^{t}=(t+\lambda+1)e^{t}(x-t)$. Requiring it to pass through $(0,0)$ and dividing by $e^{t}\neq0$:

\[ -(t+\lambda) = -t(t+\lambda+1). \tag{1} \]

Step 2: Solve for $\lambda$ in terms of $t$

Expanding and collecting the $\lambda$ terms, $\lambda(t-1)=-t^{2}$, so

\[ \lambda = \frac{t^{2}}{1-t}, \qquad t\neq1. \tag{2} \]

A tangent from the origin touches the curve at $t$ exactly when $\lambda$ has this value.

Step 3: Which values of $\lambda$ are attained?

Setting $f(t)=t^{2}/(1-t)=\lambda$ gives the quadratic $t^{2}+\lambda t-\lambda=0$, which has a real solution exactly when its discriminant is non-negative: \[\Delta = \lambda^{2}+4\lambda = \lambda(\lambda+4) \ge 0 \;\Longrightarrow\; \lambda\le-4 \ \ \text{or} \ \ \lambda\ge0. \tag{3}\]

Step 4: The excluded case $t=1$ never arises

If $t=1$ solved $t^{2}+\lambda t-\lambda=0$, it would give $1=0$ — a contradiction. So $t\neq1$ introduces no further restriction.

Step 5: How many tangents

$\Delta\gt 0$ ($\lambda\lt -4$ or $\lambda\gt 0$): two tangents. $\Delta=0$ ($\lambda=0$ or $\lambda=-4$): one (double) tangent. $-4\lt \lambda\lt 0$: no tangent from the origin.

Boundary $\lambda=0$: $t=0$; tangent $y=(0+0+1)e^{0}x=x$, through the origin $\checkmark$. Boundary $\lambda=-4$: $(t-2)^{2}=0\Rightarrow t=2$; tangent slope $(2-4+1)e^{2}=-e^{2}$, tangent $y=-e^{2}x$, through the origin $\checkmark$.

Step 6: Verification

\[\lambda(\lambda+4)\ge0 \Leftrightarrow \lambda\in(-\infty,-4]\cup[0,\infty)\ \checkmark\]

\[\lambda=-4,\ t=2\text{: tangent } y=-e^2x,\ \text{through } O\ \checkmark;\qquad \lambda=0,\ t=0\text{: tangent } y=x\ \checkmark\]