If the maximum distance of normal to the ellipse $$\dfrac{x^2}{4} + \dfrac{y^2}{b^2} = 1, b < 2$$, from the origin is 1, then the eccentricity of the ellipse is:

To find the eccentricity of the ellipse, we use the equation of the normal to the ellipse and find its maximum distance from the origin.

The given ellipse is:

$$\frac{x^2}{4} + \frac{y^2}{b^2} = 1$$

Here, the semi-major axis is $$a = 2$$ and the semi-minor axis is $$b$$ (since $$b < 2$$).

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Step 1: Write the equation of the normal to the ellipse

The equation of the normal to the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ at any parametric point $$(a\cos\theta, b\sin\theta)$$ is:

$$ax\sec\theta - by\cosec\theta = a^2 - b^2$$

Substituting $$a = 2$$ into the equation gives:

$$2x\sec\theta - by\cosec\theta = 4 - b^2$$

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Step 2: Find the distance of the normal from the origin

The perpendicular distance $$p$$ of a line $$Ax + By + C = 0$$ from the origin $$(0,0)$$ is given by the formula:

$$p = \frac{|C|}{\sqrt{A^2 + B^2}}$$

Applying this formula to our normal line:

$$p = \frac{4 - b^2}{\sqrt{4\sec^2\theta + b^2\cosec^2\theta}}$$

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Step 3: Maximize the distance from the origin

To maximize the distance $$p$$, we need to minimize the expression in the denominator:

$$f(\theta) = 4\sec^2\theta + b^2\cosec^2\theta$$

Using the identity $$\sec^2\theta = 1 + \tan^2\theta$$ and $$\cosec^2\theta = 1 + \cot^2\theta$$:

$$f(\theta) = 4(1 + \tan^2\theta) + b^2(1 + \cot^2\theta) = 4 + b^2 + 4\tan^2\theta + b^2\cot^2\theta$$

By applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality on the variable terms:

$$4\tan^2\theta + b^2\cot^2\theta \ge 2\sqrt{4\tan^2\theta \cdot b^2\cot^2\theta} = 4b$$

Thus, the minimum value of the denominator's function is:

$$f(\theta)_{\text{min}} = 4 + b^2 + 4b = (2 + b)^2$$

Substituting this back into the expression for $$p$$ gives the maximum distance:

$$p_{\text{max}} = \frac{4 - b^2}{\sqrt{(2 + b)^2}} = \frac{4 - b^2}{2 + b} = 2 - b$$

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Step 4: Solve for $$b$$ and find the eccentricity $$e$$

We are given that the maximum distance of the normal from the origin is 1:

$$2 - b = 1 \implies b = 1$$

Now, using the standard relationship for the eccentricity $$e$$ of an ellipse:

$$b^2 = a^2(1 - e^2)$$

Substituting $$a = 2$$ and $$b = 1$$:

$$1^2 = 2^2(1 - e^2)$$

$$1 = 4(1 - e^2)$$

$$1 - e^2 = \frac{1}{4} \implies e^2 = \frac{3}{4} \implies e = \frac{\sqrt{3}}{2}$$

Therefore, the eccentricity of the ellipse is equal to $$\frac{\sqrt{3}}{2}$$.