If S and S' are the foci of the ellipse $$\frac{x^2}{18} + \frac{y^2}{9} = 1$$ and P be a point on the ellipse, then $$\min(SP \cdot S'P) + \max(SP \cdot S'P)$$ is equal to:

The given ellipse is $$\frac{x^{2}}{18}+\frac{y^{2}}{9}=1$$.

Comparing with $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ we get $$a^{2}=18,\;b^{2}=9$$.
Hence $$a=3\sqrt{2},\;b=3$$.

For an ellipse, $$c^{2}=a^{2}-b^{2}$$, so
$$c^{2}=18-9=9\;\Longrightarrow\;c=3$$.

The foci are $$S(-3,0)$$ and $$S'(3,0)$$.

Let an arbitrary point on the ellipse be $$P(x,y)$$. Because $$P$$ lies on the ellipse,

$$\frac{x^{2}}{18}+\frac{y^{2}}{9}=1 \quad\Longrightarrow\quad y^{2}=9\left(1-\frac{x^{2}}{18}\right)=9-\frac{x^{2}}{2}\;.$$ $$-(1)$$

Distances of $$P$$ from the two foci are

$$SP=\sqrt{(x+3)^{2}+y^{2}},\qquad S'P=\sqrt{(x-3)^{2}+y^{2}}.$$

We need the product $$SP\cdot S'P$$. First square each distance:

$$SP^{2}=(x+3)^{2}+y^{2},\qquad S'P^{2}=(x-3)^{2}+y^{2}.$$

Using $$(1)$$ to eliminate $$y^{2}$$:

$$SP^{2}=(x+3)^{2}+9-\frac{x^{2}}{2}$$ $$\phantom{SP^{2}}=\frac{x^{2}}{2}+6x+9+9 =\frac{x^{2}}{2}+6x+18,$$

$$S'P^{2}=(x-3)^{2}+9-\frac{x^{2}}{2}$$ $$\phantom{S'P^{2}}=\frac{x^{2}}{2}-6x+18.$$

Therefore

$$\bigl(SP\cdot S'P\bigr)^{2}=SP^{2}\,S'P^{2} =\left(\frac{x^{2}}{2}+6x+18\right)\left(\frac{x^{2}}{2}-6x+18\right).$$

Recognise the form $$(A+6x)(A-6x)=A^{2}-36x^{2},$$ where $$A=\dfrac{x^{2}}{2}+18$$. Hence

$$\bigl(SP\cdot S'P\bigr)^{2} =\left(\frac{x^{2}}{2}+18\right)^{2}-36x^{2}.$$ Expand: $$\bigl(SP\cdot S'P\bigr)^{2} =\frac{x^{4}}{4}+18x^{2}+324-36x^{2} =\frac{x^{4}}{4}-18x^{2}+324.$$ Let $$X=x^{2}\;(X\ge 0)$$ to work with one variable: $$\bigl(SP\cdot S'P\bigr)^{2} =\frac{X^{2}}{4}-18X+324.$$ $$-(2)$$

The point $$P$$ lies on the ellipse, so $$x^{2}\le a^{2}=18$$, that is $$0\le X\le 18$$.

Equation $$(2)$$ is a quadratic in $$X$$ with coefficient $$\frac14>0$$; it opens upward. Its vertex is at

$$X_v=\frac{-(-18)}{2\left(\frac14\right)} =\frac{18}{0.5}=36.$$

The vertex lies to the right of the allowed interval $$[0,18]$$, so within the interval the quadratic is strictly decreasing.
Hence:

• Maximum of $$(SP\cdot S'P)^{2}$$ occurs at the left end $$X=0$$.
• Minimum of $$(SP\cdot S'P)^{2}$$ occurs at the right end $$X=18$$.

From $$(2):\;\bigl(SP\cdot S'P\bigr)^{2}=324 \;\Longrightarrow\;SP\cdot S'P=18.$$

Substitute $$X=18$$ in $$(2):$$ $$\bigl(SP\cdot S'P\bigr)^{2} =\frac{(18)^{2}}{4}-18(18)+324 =81-324+324=81,$$ $$\Longrightarrow\;SP\cdot S'P=9.$$

Thus

$$\min(SP\cdot S'P)=9,\qquad \max(SP\cdot S'P)=18.$$

Required sum:

$$\min(SP\cdot S'P)+\max(SP\cdot S'P)=9+18=27.$$

Option D is correct.