If the normal to the ellipse $$3x^2 + 4y^2 = 12$$ at a point P on it is parallel to the line, $$2x + y = 4$$ and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to:

We have the ellipse $$3x^{2}+4y^{2}=12$$ and a point $$P(x_{1},y_{1})$$ on it.

The line $$2x+y=4$$ is given to be parallel to the normal at $$P$$. The slope of this line is obtained by rewriting it in slope-intercept form:

$$y=-2x+4 \;\;\Longrightarrow\;\; \text{slope} = -2.$$

Recall the standard differentiation result: for any curve $$F(x,y)=0,$$ differentiating gives $$F_{x}+F_{y}\dfrac{dy}{dx}=0\quad\Longrightarrow\quad \dfrac{dy}{dx}=-\dfrac{F_{x}}{F_{y}}.$$ This $$\dfrac{dy}{dx}$$ is the slope of the tangent. The slope of the normal is its negative reciprocal.

For our ellipse,

$$F(x,y)=3x^{2}+4y^{2}-12=0,$$ so $$F_{x}=6x,\qquad F_{y}=8y.$$

Hence

$$\dfrac{dy}{dx}=-\dfrac{6x}{8y}=-\dfrac{3x}{4y} \quad\Longrightarrow\quad \text{Slope of normal }m_{n}=-\dfrac{1}{m_{t}}=\dfrac{4y}{3x}.$$

The normal is parallel to the line whose slope is $$-2,$$ therefore

$$\dfrac{4y_{1}}{3x_{1}}=-2 \;\;\Longrightarrow\;\; 4y_{1}=-6x_{1} \;\;\Longrightarrow\;\; x_{1}=-\dfrac{2}{3}y_{1}. \quad -(1)$$

Point $$P(x_{1},y_{1})$$ also lies on the ellipse, so

$$3x_{1}^{2}+4y_{1}^{2}=12.$$

Substituting the relation (1):

$$3\left(-\dfrac{2}{3}y_{1}\right)^{2}+4y_{1}^{2}=12$$ $$\Longrightarrow\; 3\left(\dfrac{4}{9}y_{1}^{2}\right)+4y_{1}^{2}=12$$ $$\Longrightarrow\; \dfrac{12}{9}y_{1}^{2}+4y_{1}^{2}=12$$ $$\Longrightarrow\; \dfrac{4}{3}y_{1}^{2}+4y_{1}^{2}=12$$ $$\Longrightarrow\; \dfrac{4+12}{3}y_{1}^{2}=12$$ $$\Longrightarrow\; \dfrac{16}{3}y_{1}^{2}=12$$ $$\Longrightarrow\; y_{1}^{2}=12\cdot\dfrac{3}{16}=\dfrac{36}{16}=\dfrac{9}{4}.$$

So

$$y_{1}=\pm\dfrac{3}{2},\qquad x_{1}=-\dfrac{2}{3}y_{1}\;\;\Longrightarrow\;\; x_{1}=\mp1.$$

The two candidate points are $$P_{1}\,(-1,\;3/2)\quad\text{and}\quad P_{2}\,(1,\;-3/2).$$

Next, the tangent at $$P$$ must pass through $$Q(4,4).$$ For the ellipse $$Ax^{2}+By^{2}=C,$$ the tangent at $$(x_{1},y_{1})$$ is $$Axx_{1}+Byy_{1}=C.$$ Here $$A=3,\;B=4,\;C=12,$$ so the tangent at $$P(x_{1},y_{1})$$ is

$$3xx_{1}+4yy_{1}=12.$$

Substituting $$Q(4,4)$$ gives the condition

$$3\cdot4\,x_{1}+4\cdot4\,y_{1}=12 \;\;\Longrightarrow\;\; 12x_{1}+16y_{1}=12 \;\;\Longrightarrow\;\; 3x_{1}+4y_{1}=3. \quad -(2)$$

We test both points:

For $$P_{1}(-1,3/2):$$ $$3(-1)+4\left(\dfrac{3}{2}\right)=-3+6=3,$$ so equation (2) is satisfied.

For $$P_{2}(1,-3/2):$$ $$3(1)+4\left(-\dfrac{3}{2}\right)=3-6=-3\neq3,$$ so $$P_{2}$$ is rejected.

Therefore $$P(-1,\;3/2).$$

Finally, the distance $$PQ$$ where $$Q(4,4)$$ is

$$PQ=\sqrt{(4-(-1))^{2}+\left(4-\dfrac{3}{2}\right)^{2}} =\sqrt{5^{2}+\left(\dfrac{5}{2}\right)^{2}} =\sqrt{25+\dfrac{25}{4}} =\sqrt{\dfrac{125}{4}} =\dfrac{\sqrt{125}}{2} =\dfrac{5\sqrt{5}}{2}.$$

Hence, the correct answer is Option B.