If the tangent to the curve $$y = e^x$$ at a point $$(c, e^c)$$ and the normal to the parabola $$y^2 = 4x$$ at the point (1, 2) intersect at the same point on the $$x$$-axis, then the value of $$c$$ is .....

Let us first consider the curve $$y = e^x$$. The point of tangency is given to be $$(c,\,e^{\,c})$$. For any curve $$y=f(x)$$, the slope of the tangent at a point $$x=c$$ is obtained from the derivative $$\dfrac{dy}{dx}=f'(x)$$ evaluated at $$x=c$$. Here $$f(x)=e^{x}$$, and we know the standard result $$\dfrac{d}{dx}\bigl(e^{x}\bigr)=e^{x}$$. Therefore at $$x=c$$ the slope of the tangent is $$e^{\,c}$$.

Using the point-slope form of a straight line, $$y-y_{1}=m\,(x-x_{1})$$, the equation of the tangent through the point $$(c,\,e^{\,c})$$ with slope $$m=e^{\,c}$$ is

$$y-e^{\,c}=e^{\,c}\,(x-c).$$

To locate the point where this tangent meets the $$x$$-axis we set $$y=0$$, because every point on the $$x$$-axis has ordinate zero. Substituting $$y=0$$ in the tangent equation we have

$$0-e^{\,c}=e^{\,c}\,(x-c).$$

Simplifying,

$$-e^{\,c}=e^{\,c}\,(x-c).$$

Dividing both sides by the non-zero quantity $$e^{\,c}$$ yields

$$-1=x-c.$$

So the $$x$$-coordinate of the intersection is

$$x=c-1.$$

Hence the tangent to $$y=e^{x}$$ meets the $$x$$-axis at the point $$\bigl(c-1,\,0\bigr).$$

Now we turn to the parabola $$y^{2}=4x$$ and look at its normal at the point $$(1,\,2)$$. First we need the derivative of the parabola to obtain the slope of the tangent. Differentiating implicitly,

$$2y\,\dfrac{dy}{dx}=4,$$

so

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

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

$$m_{\text{tangent}}=\frac{2}{2}=1.$$

The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore

$$m_{\text{normal}}=-\frac{1}{1}=-1.$$

Again using the point-slope form, the equation of the normal through $$(1,\,2)$$ with slope $$-1$$ is

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

To find where this normal meets the $$x$$-axis we put $$y=0$$:

$$0-2=-1\,(x-1).$$

Thus

$$-2=-x+1.$$

Multiplying by $$-1$$ to make the algebra clearer,

$$2=x-1,$$

and therefore

$$x=3.$$

Consequently, the normal to the parabola meets the $$x$$-axis at the point $$(3,\,0).$$

The statement of the problem tells us that the tangent to $$y=e^{x}$$ and the normal to $$y^{2}=4x$$ intersect the $$x$$-axis at the same point. We have found the $$x$$-coordinate of that common point in two ways: the tangent gives $$x=c-1$$, and the normal gives $$x=3$$. Equating these:

$$c-1 = 3.$$

Adding $$1$$ to both sides, we obtain

$$c = 4.$$

Hence, the correct answer is Option 4.