How to find the value of "k" in following condition?

Question

For what values of $k$ the line $y=9x$ be the tangent to the curve $y=\frac{k e^{k \sqrt{x+1}}}{\sqrt{x+1}}$ at some point on the $xy-$plane with constraint that $x>-1$

options are given:

a) k>0
b) k>0 and k<1
c) k>1 and k<3
d) k>3

this can be done like by using condition as $$\begin{align} &\frac{dy}{dx}=9\\ &= \frac{k e^{k \sqrt{x+1}}}{2\sqrt{x+1}}\left(\frac{k \sqrt{x+1}-1}{x+1} \right)\\ \end{align}$$

what to do next to find the value of k, how to use the given constraint does this use for only sqrt to be defined, thanks.

Answer 1

In fact, the curve $$\mathcal{T}=\left\{(x,y_T)\in \mathbb{R}^2: y_S=9x \right\}$$ must be the tangent of the curve $$\mathcal{S}_k=\left\{(x,y_S)\in \mathbb{R}^2: y_S= \frac{k e^{k\sqrt{x+1}}}{\sqrt{x+1}}, x>-1 \right\}.$$

So, for this to be true, we need that \begin{align} \exists (x^*,y_S^*)\in \mathcal{S} \text{ and } (x^*,y_T^*) \in \mathcal{T} \; \text{s.t.} \; \frac{\mathrm{d}}{\mathrm{d} x} y_T \Big{|}_{(x^*,y_T^*)} = \frac{\mathrm{d}}{\mathrm{d} x} y_S \Big{|}_{(x^*,y_S^*)} \; \text{and} \; (x^*,y_T^*) = (x^*,y_S^*). \end{align}

From there you should be able concluded your answer. As your equations look suspicious, I let the raw method here.

Answer 2

(too long for a comment)

Let $(s,t)$ be the tangent point.

We have $$y'=\frac{ke^{k\sqrt{x+1}}(k\sqrt{x+1}-1)}{2(x+1)^{3/2}}$$ so we want $k$ such that there exists $(s,t)$ satisfying $$t=9s$$ $$t=\frac{ke^{k\sqrt{s+1}}}{\sqrt{s+1}}$$ $$\frac{ke^{k\sqrt{s+1}}(k\sqrt{s+1}-1)}{2(s+1)^{3/2}}=9$$ From these, we can represent $k$ by $s$ and have an equation on $s$, but solving it seems difficult.