The tangent to the parabola $$y^{2} = 4x$$ at the point where it intersects the circle $$x^{2} + y^{2} = 5$$ in the first quadrant, passes through the point:

We begin with the two curves that have been given.

The parabola is $$y^{2}=4x$$ and the circle is $$x^{2}+y^{2}=5.$$

We want the common point of these two curves that lies in the first quadrant, that is, with $$x>0$$ and $$y>0.$$

For a parabola of the form $$y^{2}=4ax,$$ a convenient parametric representation is

$$x = at^{2},\qquad y = 2at.$$

Here the parameter is $$t$$ and the constant $$a=1$$ (because the right side is $$4x = 4\cdot1\cdot x$$). Hence for our parabola we set

$$x = t^{2},\qquad y = 2t.$$

Now we substitute these expressions for $$x$$ and $$y$$ in the equation of the circle $$x^{2}+y^{2}=5.$$ So we have

$$\left(t^{2}\right)^{2} + \left(2t\right)^{2} = 5.$$

Simplifying each term gives

$$t^{4} + 4t^{2} = 5.$$

To solve this quartic, let us set $$u = t^{2}.$$ Because $$t^{2} \ge 0,$$ we will look only for non-negative solutions of $$u.$$ Rewriting, we get

$$u^{2} + 4u - 5 = 0.$$

This is a quadratic in $$u.$$ Using the quadratic formula $$u=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a},$$ with $$a=1,\; b=4,\; c=-5,$$ we find

$$u = \dfrac{-4 \pm \sqrt{16 + 20}}{2} = \dfrac{-4 \pm \sqrt{36}}{2} = \dfrac{-4 \pm 6}{2}.$$

That yields two numerical values:

$$u = \dfrac{-4 + 6}{2} = 1, \qquad u = \dfrac{-4 - 6}{2} = -5.$$

Because $$u = t^{2} \ge 0,$$ we discard the negative value $$u=-5$$ and keep $$u=1.$$

Thus $$t^{2}=1 \implies t=\pm1.$$ We need the point in the first quadrant where $$y=2t$$ is positive, so we choose $$t=+1.$$

With $$t=1$$ we obtain the coordinates of the point of intersection: $$x = t^{2} = 1,\qquad y = 2t = 2.$$ Therefore the required point on both the parabola and the circle in the first quadrant is $$P(1,2).$$

Next, we find the equation of the tangent to the parabola at the point $$P.$$

First, we recall that for any curve given implicitly as $$F(x,y)=0,$$ the slope of the tangent is obtained via implicit differentiation: $$\frac{dy}{dx} = -\frac{F_{x}}{F_{y}}.$$

For the parabola $$y^{2}=4x,$$ let us differentiate both sides with respect to $$x.$$ We have

$$\frac{d}{dx}\bigl(y^{2}\bigr) = \frac{d}{dx}\bigl(4x\bigr).$$

Using the power rule and the chain rule,

$$2y\frac{dy}{dx} = 4.$$

Hence

$$\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}.$$

At the point $$P(1,2),$$ we substitute $$y=2$$ to get the slope of the tangent:

$$m = \frac{dy}{dx}\Big|_{(1,2)} = \frac{2}{2} = 1.$$

Now we use the point-slope form of a straight line:

$$y - y_{1} = m(x - x_{1}).$$

With $$m = 1,\; (x_{1},y_{1}) = (1,2),$$ this becomes

$$y - 2 = 1\,(x - 1).$$

Simplifying gives

$$y - 2 = x - 1 \;\;\Longrightarrow\;\; y = x + 1.$$

This is the equation of the tangent line to the parabola at the required point.

Finally, we check which of the four option points lies on the line $$y = x + 1.$$ For a point $$\bigl(x_{0},y_{0}\bigr)$$ to be on this line we must have $$y_{0}=x_{0}+1.$$

Let us test each option:

Option A $$\left(\frac14,\frac34\right):$$ $$\frac34 \neq \frac14 + 1 = \frac54,$$ so it is not on the line.

Option B $$\left(-\frac13,\frac43\right):$$ $$\frac43 \neq -\frac13 + 1 = \frac23,$$ so it fails as well.

Option C $$\left(-\frac14,\frac12\right):$$ $$\frac12 \neq -\frac14 + 1 = \frac34,$$ so it is not on the line.

Option D $$\left(\frac34,\frac74\right):$$ Here $$\frac74 = \frac34 + 1,$$ which is exactly satisfied.

Therefore the tangent line $$y = x + 1$$ passes through the point $$\left(\frac34,\frac74\right),$$ which corresponds to Option 4.

Hence, the correct answer is Option 4.