Question:

Step 1: Identify the common center of the ellipse and hyperbola. The given equation for the ellipse is $E : \dfrac{(x-1)^2}{16} + \dfrac{(y-2)^2}{9} = 1$. The given equation for the hyperbola is $H : (x-1)^2 - (y-2)^2 = \dfrac{7}{2}$. Both equations are of the form $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$ and $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$ respectively, where $(h,k)$ is the center. For both curves, the center is $(1, 2)$. Step 2: Express $(x-1)^2$ in terms of $(y-2)^2$ from the hyperbola equation. From the equation of the hyperbola $H$: $$(x-1)^2 - (y-2)^2 = \frac{7}{2}$$ Rearranging to isolate $(x-1)^2$, we get: $$(x-1)^2 = (y-2)^2 + \frac{7}{2}$$ Step 3: Substitute the expression for $(x-1)^2$ into the ellipse equation. Substitute $(x-1)^2 = (y-2)^2 + \frac{7}{2}$ into the equation of the ellipse $E$: $$\frac{(y-2)^2 + \frac{7}{2}}{16} + \frac{(y-2)^2}{9} = 1$$ Step 4: Solve the resulting equation for $(y-2)^2$. To eliminate the denominators, multiply the entire equation by the least common multiple of 16 and 9, which is 144: $$144 \left( \frac{(y-2)^2 + \frac{7}{2}}{16} + \frac{(y-2)^2}{9} \right) = 144 \times 1$$ $$9 \left( (y-2)^2 + \frac{7}{2} \right) + 16 (y-2)^2 = 144$$ Expand the terms: $$9(y-2)^2 + \frac{63}{2} + 16(y-2)^2 = 144$$ Combine the terms involving $(y-2)^2$: $$25(y-2)^2 + 31.5 = 144$$ Subtract 31.5 from both sides: $$25(y-2)^2 = 144 - 31.5$$ $$25(y-2)^2 = 112.5$$ Divide by 25: $$(y-2)^2 = \frac{112.5}{25}$$ $$(y-2)^2 = 4.5 = \frac{9}{2}$$ Taking the square root of both sides: $$y-2 = \pm\sqrt{\frac{9}{2}} = \pm\frac{3}{\sqrt{2}}$$ Step 5: Determine the values for $(x-1)^2$. Substitute $(y-2)^2 = \frac{9}{2}$ back into the expression for $(x-1)^2$ from Step 2: $$(x-1)^2 = (y-2)^2 + \frac{7}{2}$$ $$(x-1)^2 = \frac{9}{2} + \frac{7}{2}$$ $$(x-1)^2 = \frac{16}{2}$$ $$(x-1)^2 = 8$$ Taking the square root of both sides: $$x-1 = \pm\sqrt{8} = \pm 2\sqrt{2}$$ Step 6: Determine the number of intersection points and match with the given options. From $y-2 = \pm\frac{3}{\sqrt{2}}$, we have two distinct values for $y$. From $x-1 = \pm 2\sqrt{2}$, we have two distinct values for $x$. Combining these, there are $2 \times 2 = 4$ distinct points of intersection between the ellipse and the hyperbola. The coordinates of these intersection points are $(1 \pm 2\sqrt{2}, 2 \pm \frac{3}{\sqrt{2}})$. The existence of four distinct intersection points means that option (d) (IV) (ii) (R) is the correct combination. The final answer is $\boxed{\text{D}}$.