Answer
the value of the number a is $\sqrt[3]{3}$
Work Step by Step
Let
$$
y=(x+c)^{-1}
$$
$\Rightarrow
$
$$
y^{\prime}=-(x+c)^{-2} $$
and
$$ y=a(x+k)^{1 / 2}$$
$ \Rightarrow$
$$
y^{\prime}=\frac{1}{3} a(x+k)^{-2 / 3}$$
so the curves are orthogonal if the product of the slopes is -1, that is,
$$\frac{-1}{(x+c)^{2}} \cdot \frac{a}{3(x+k)^{2 / 3}}=-1 $$
$
\Rightarrow
$
$$
\begin{aligned}
a & =3(x+c)^{2}(x+k)^{2 / 2}\\
& \ \ \ \ \left[\text { since } y^{2}=(x+c)^{-2} \text {and } y^{2}=a^{2}(x+k)^{2 / 3}\right]\\
&=3\left(\frac{1}{y}\right)^{2}\left(\frac{y}{a}\right)^{2}\\
\end{aligned}
$$
$ \Rightarrow $
$$
a=3\left(\frac{1}{a^{2}}\right) \Rightarrow a^{3}=3 \Rightarrow a=\sqrt[3]{3}$$.
Hence the value of the number a is $\sqrt[3]{3}$
