Key Concepts and Formulas

To solve this problem, we will utilize the following fundamental concepts related to ellipses:

1. Equation of the Normal to an Ellipse: For an ellipse given by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of the normal at a point $P(a \cos \theta, b \sin \theta)$ (in parametric form) is given by: $ax \sec \theta - by \csc \theta = a^2 - b^2$ Here, $a$ and $b$ represent the lengths of the semi-major and semi-minor axes, respectively.

2. Perpendicular Distance from a Point to a Line: The perpendicular distance $D$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by: $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

3. Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For non-negative real numbers $X$ and $Y$, the AM-GM inequality states that $\frac{X+Y}{2} \ge \sqrt{XY}$. Equality holds if and only if $X=Y$. This inequality is particularly useful for finding minimum or maximum values of expressions.

4. Eccentricity of an Ellipse: For an ellipse with semi-major axis $a$ and semi-minor axis $b$, the eccentricity $e$ is related by the formula: $b^2 = a^2(1-e^2)$

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Step-by-Step Solution

1. Identify Ellipse Parameters and Set Up the Equation of the Normal

• Understanding the given ellipse: The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{b^2} = 1$.
  • Comparing this with the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, we have $A^2 = 4$ and $B^2 = b^2$.
  • This means the squares of the semi-axes are $4$ and $b^2$.
• Determining semi-major and semi-minor axes: The problem states $b < 2$. Since $A=\sqrt{4}=2$, this implies that $A=2$ is the length of the semi-major axis, and $B=b$ is the length of the semi-minor axis.
  • Therefore, in our standard formulas, we will use $a=2$ (for the semi-major axis) and $b$ (for the semi-minor axis, as given in the problem).
• Equation of the Normal: Substitute $a=2$ into the general equation of the normal: $2x \sec \theta - by \csc \theta = 2^2 - b^2$ $2x \sec \theta - by \csc \theta = 4 - b^2 \quad (*)$

2. Calculate the Perpendicular Distance of the Normal from the Origin

• Goal: We need to find the distance of the normal line $(*)$ from the origin $(0,0)$.
• Applying the distance formula: For the line $2x \sec \theta - by \csc \theta - (4 - b^2) = 0$, the coefficients are $A = 2 \sec \theta$, $B = -b \csc \theta$, and $C = -(4 - b^2)$. The point is $(x_0, y_0) = (0,0)$. $D = \frac{|(2 \sec \theta)(0) + (-b \csc \theta)(0) - (4 - b^2)|}{\sqrt{(2 \sec \theta)^2 + (-b \csc \theta)^2}}$ $D = \frac{|-(4 - b^2)|}{\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}}$ $D = \frac{|4 - b^2|}{\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}}$
• Simplifying the numerator: Since $b < 2$, $b^2 < 4$. Therefore, $4 - b^2$ is positive, and $|4 - b^2| = 4 - b^2$. $D = \frac{4 - b^2}{\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}}$

3. Maximize the Distance $D$

• Condition: The problem states that the maximum distance of the normal from the origin is 1.
• Strategy for maximization: To maximize $D = \frac{4 - b^2}{\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}}$, since the numerator $(4-b^2)$ is a positive constant, we need to minimize the denominator $\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}$. This is equivalent to minimizing the expression inside the square root: $f(\theta) = 4 \sec^2 \theta + b^2 \csc^2 \theta$.
• Using trigonometric identities: Recall that $\sec^2 \theta = 1 + \tan^2 \theta$ and $\csc^2 \theta = 1 + \cot^2 \theta$. $f(\theta) = 4(1 + \tan^2 \theta) + b^2(1 + \cot^2 \theta)$ $f(\theta) = 4 + 4 \tan^2 \theta + b^2 + b^2 \cot^2 \theta$ $f(\theta) = (4 + b^2) + (4 \tan^2 \theta + b^2 \cot^2 \theta)$
• Applying AM-GM inequality: Let $X = 4 \tan^2 \theta$ and $Y = b^2 \cot^2 \theta$. Since $\tan^2 \theta$ and $\cot^2 \theta$ are non-negative, $X$ and $Y$ are non-negative. $X + Y \ge 2\sqrt{XY}$ $4 \tan^2 \theta + b^2 \cot^2 \theta \ge 2\sqrt{(4 \tan^2 \theta)(b^2 \cot^2 \theta)}$ $4 \tan^2 \theta + b^2 \cot^2 \theta \ge 2\sqrt{4b^2 \tan^2 \theta \cot^2 \theta}$ $4 \tan^2 \theta + b^2 \cot^2 \theta \ge 2\sqrt{4b^2 \cdot 1} \quad (\text{since } \tan \theta \cot \theta = 1)$ $4 \tan^2 \theta + b^2 \cot^2 \theta \ge 2(2b)$ $4 \tan^2 \theta + b^2 \cot^2 \theta \ge 4b$
• Minimum value of the denominator term: Substituting this back into $f(\theta)$: $f(\theta)_{min} = (4 + b^2) + 4b = (2+b)^2$ This minimum occurs when $4 \tan^2 \theta = b^2 \cot^2 \theta$, which implies $\tan^4 \theta = b^2/4$, or $\tan^2 \theta = b/2$. Such a $\theta$ always exists.
• Minimum value of the square root: $$$$