On the ellipse $$\frac{x^2}{8} + \frac{y^2}{4} = 1$$, let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line $$x + 2y = 0$$. Let S and S' be the foci of the ellipse and $$e$$ be its eccentricity. If A is the area of the triangle SPS', then the value of $$(5 - e^2) \cdot A$$ is

We start with the ellipse $$\dfrac{x^{2}}{8}+\dfrac{y^{2}}{4}=1$$ whose semi-major axis is along the x-axis because $$a^{2}=8\gt b^{2}=4$$. So we have $$a=\sqrt 8=2\sqrt2,\;b=\sqrt4=2.$$

The given condition is that the tangent at a point $$P(x_{1},y_{1})$$ on the ellipse is perpendicular to the straight line $$x+2y=0.$$

First, write the slope of the given straight line. Solving $$x+2y=0$$ for $$y$$ gives $$y=-\dfrac{x}{2},$$ so its slope is $$m_{1}=-\dfrac12.$$

If two lines are perpendicular, then the product of their slopes equals $$-1.$$ Hence, if the slope of the tangent at $$P$$ is $$m,$$ we must have

$$m\cdot\Bigl(-\dfrac12\Bigr)=-1\;\Longrightarrow\;m=2.$$

Now, we need the slope of the tangent to the ellipse. Using implicit differentiation on $$\dfrac{x^{2}}{8}+\dfrac{y^{2}}{4}=1,$$ we have

$$$\dfrac{2x}{8}+\dfrac{2y}{4}\dfrac{dy}{dx}=0 \;\Longrightarrow\;\dfrac{x}{4}+ \dfrac{y}{2}\dfrac{dy}{dx}=0 \;\Longrightarrow\;\dfrac{dy}{dx}=-\dfrac{x}{2y}.$$$

Therefore the slope at $$P(x_{1},y_{1})$$ is $$m=-\dfrac{x_{1}}{2y_{1}}.$$ Equating this slope to the required value $$2,$$ we get

$$-\dfrac{x_{1}}{2y_{1}}=2 \;\Longrightarrow\;x_{1}=-4y_{1}.$$

Because $$P$$ lies on the ellipse, its coordinates must also satisfy $$\dfrac{x_{1}^{2}}{8}+\dfrac{y_{1}^{2}}{4}=1.$$ Substituting $$x_{1}=-4y_{1}$$ into this equation, we obtain

$$$\dfrac{(-4y_{1})^{2}}{8}+\dfrac{y_{1}^{2}}{4}=1 \;\Longrightarrow\;\dfrac{16y_{1}^{2}}{8}+\dfrac{y_{1}^{2}}{4}=1 \;\Longrightarrow\;2y_{1}^{2}+\dfrac{y_{1}^{2}}{4}=1 \;\Longrightarrow\;\dfrac{8y_{1}^{2}+y_{1}^{2}}{4}=1 \;\Longrightarrow\;\dfrac{9y_{1}^{2}}{4}=1 \;\Longrightarrow\;y_{1}^{2}=\dfrac{4}{9} \;\Longrightarrow\;y_{1}=\dfrac{2}{3}\quad(\text{positive because }P\text{ is in the second quadrant}).$$$

Using $$x_{1}=-4y_{1},$$ we now get

$$x_{1}=-4\left(\dfrac{2}{3}\right)=-\dfrac{8}{3}.$$

Thus the required point is $$P\!\left(-\dfrac83,\dfrac23\right).$$ Its x-coordinate is negative and y-coordinate positive, confirming that $$P$$ is indeed in the second quadrant.

Next we determine the foci of the ellipse. For an ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ with $$a\gt b,$$ the distance of each focus from the centre is $$c=\sqrt{a^{2}-b^{2}}.$$ Here

$$c=\sqrt{8-4}=2.$$

Hence the foci are $$S(2,0)$$ and $$S'(-2,0).$$ The eccentricity $$e$$ is defined as $$e=\dfrac{c}{a},$$ so

$$e=\dfrac{2}{2\sqrt2}=\dfrac1{\sqrt2},\qquad e^{2}=\dfrac12.$$

We must now find the area $$A$$ of the triangle $$\triangle SPS'$$ with vertices

$$$S(2,0),\;P\!\left(-\dfrac83,\dfrac23\right),\;S'(-2,0).$$$

The simplest approach is the coordinate (shoelace) formula for the area of a triangle:

$$$A=\dfrac12\left|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right|.$$$

Label $$$S(x_{1},y_{1})=(2,0),\;P(x_{2},y_{2})=\left(-\dfrac83,\dfrac23\right),\;S'(x_{3},y_{3})=(-2,0).$$$ Substituting, we get

$$$ \begin{aligned} A&=\dfrac12\Biggl|\,2\!\left(\dfrac23-0\right) +\left(-\dfrac83\right)(0-0) +(-2)\!\left(0-\dfrac23\right)\Biggr|\\[2mm] &=\dfrac12\Biggl|\;2\cdot\dfrac23+0+(-2)\!\left(-\dfrac23\right)\Biggr|\\[2mm] &=\dfrac12\Biggl|\;\dfrac43+\dfrac43\Biggr|\\[2mm] &=\dfrac12\cdot\dfrac83=\dfrac43. \end{aligned} $$$

So, $$A=\dfrac43.$$

Finally we evaluate $$(5-e^{2})\cdot A.$$ We have already found $$e^{2}=\dfrac12,$$ therefore

$$$ (5-e^{2})\cdot A =\left(5-\dfrac12\right)\cdot\dfrac43 =\dfrac{9}{2}\cdot\dfrac43 =\dfrac{9\cdot4}{2\cdot3} =\dfrac{36}{6} =6. $$$

Hence, the correct answer is Option B.