Edited (Not the slope, but the orthogonal direction)
The correct is the one from your textbook. In polars, the perpendicular is got by means $\dfrac{1}{r}\dfrac{\mathrm dr}{\mathrm d\theta}\to-r\dfrac{\mathrm d\theta}{\mathrm dr}$
Added
Suppose the position vector for each curve points with parameter $t$: $(r,\theta)=(r(t),\theta(t))$
It's known the vector tangent to this curve is:
$$\frac{\mathrm d\mathbf v}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat r}+\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}$$
An orthogonal vector to this one is:
$$\mathbf v_p=r\frac{\mathrm d\mathbf \theta}{\mathrm dt}\mathbf{\hat r}-\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat\theta}$$
You can now consider eliminating the parameter for the curve:
$$\frac{\mathrm dr/\mathrm dt}{r\mathrm d\mathbf \theta/\mathrm dt}=\frac{\mathrm dr}{r\mathrm d\theta}=f(r,\theta)$$
Now,
$$-\frac{r\mathrm d\theta}{\mathrm dr}=f(r,\theta)$$
is true of the orthogonal curves to the original one.
$r=2C\cos\theta$, $t=\theta$, $(r(t),\theta(t))=(2C\cos(t),t)$
$$\frac{\mathrm d\mathbf v}{\mathrm dt}=-2C\sin(t)\mathbf{\hat r}+1·2C\cos(t)\mathbf{\hat\theta}$$$$\frac{\mathrm d\mathbf v}{\mathrm dt}=\frac{\mathrm dr}{\mathrm dt}\mathbf{\hat r}+\frac{\mathrm d\mathbf\theta}{\mathrm dt}r\mathbf{\hat\theta}=-2C\sin(t)\mathbf{\hat r}+1·2C\cos(t)\mathbf{\hat\theta}$$
$$\frac{\mathrm dr}{r\mathrm d\theta}=\frac{-2C\sin(t)}{2C\cos(t)}=\frac{-\sin\theta}{\cos\theta}$$$$\frac{\mathrm dr/\mathrm dt}{r\mathrm d\mathbf \theta/\mathrm dt}=\frac{-2C\sin(t)}{2C\cos(t)}=\frac{-\sin\theta}{\cos\theta}=\frac{\mathrm dr}{r\mathrm d\theta}$$
and
$$\frac{r\mathrm d\theta}{\mathrm dr}=\frac{\sin\theta}{\cos\theta}$$