Let C be a curve given by $$y(x) = 1 + \sqrt{4x - 3}$$, $$x > \frac{3}{4}$$. If $$P$$ is a point on C, such that the tangent at $$P$$ has slope $$\frac{2}{3}$$, then a point through which the normal at $$P$$ passes, is:

We are given the curve $$y(x)=1+\sqrt{4x-3}$$ with the condition $$x>\dfrac34$$, so the square-root remains real. A point $$P(x_P,y_P)$$ on this curve is such that the tangent at $$P$$ has slope $$\dfrac23$$. Our task is to find a point among the options through which the normal at $$P$$ passes.

First we differentiate the given function. We recall the derivative formula for a square root:

For $$u(x)$$, $$\dfrac{d}{dx}\sqrt{u}= \dfrac1{2\sqrt{u}}\dfrac{du}{dx}.$$

Here $$u(x)=4x-3$$, so $$\dfrac{du}{dx}=4$$. Therefore

$$\frac{dy}{dx}= \frac{1}{2\sqrt{4x-3}}\cdot 4 = \frac{2}{\sqrt{4x-3}}.$$

The slope of the tangent at point $$P$$ must equal the given value $$\dfrac23$$. Hence we set

$$\frac{2}{\sqrt{4x_P-3}}=\frac23.$$

We solve this equation step by step:

First cancel the common factor $$2$$ on both sides:

$$\frac{1}{\sqrt{4x_P-3}}=\frac13.$$

Now take reciprocals:

$$\sqrt{4x_P-3}=3.$$

Square both sides to remove the square root:

$$4x_P-3 = 9.$$

Add $$3$$ to both sides:

$$4x_P = 12.$$

Divide by $$4$$:

$$x_P = 3.$$

We substitute this $$x_P$$ back into the original curve to find $$y_P$$:

$$y_P = 1+\sqrt{4\cdot3-3}=1+\sqrt{12-3}=1+\sqrt9=1+3=4.$$

Hence the coordinates of the desired point on the curve are $$P(3,4).$$

The slope of the tangent at $$P$$ is $$\dfrac23$$, so the slope of the normal, being the negative reciprocal, is given by

$$m_{\text{normal}} = -\frac{1}{\left(\dfrac23\right)} = -\frac32.$$

Using the point-slope form of a straight line, the equation of the normal through $$P(3,4)$$ is

$$y-4 = -\frac32\,(x-3).$$

We now check which option satisfies this equation.

Option A: $$(1,7).$$ Substituting, $$y-4 = 7-4 = 3,$$ and $$-\frac32\,(x-3)= -\frac32\,(1-3)= -\frac32\,(-2)=3.$$ Since the left and right sides are equal, the point $$(1,7)$$ lies on the normal.

Option B: $$(3,-4).$$ Here $$x-3=0,$$ so the right side is $$0,$$ while $$y-4=-8,$$ giving a mismatch. Hence it is not on the normal.

Option C: $$(4,-3).$$ We get $$y-4=-7,$$ and $$-\dfrac32\,(4-3)=-\dfrac32,$$ which are unequal. So this point is not on the normal.

Option D: $$(2,3).$$ We obtain $$y-4=-1,$$ and $$-\dfrac32\,(2-3)= -\dfrac32\,(-1)=\dfrac32,$$ again unequal. Thus this point is not on the normal.

The only point that satisfies the normal equation is $$(1,7).$$

Hence, the correct answer is Option A.