This problem is a classic example of applying analytical geometry concepts to an ellipse. It requires finding the equation of a chord given its midpoint and then calculating its length. The most efficient approach involves using a specialized formula for the chord's equation, followed by standard algebraic techniques to determine its length.

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Key Concept: Equation of a Chord with a Given Midpoint

One of the most powerful and frequently used formulas in the study of conic sections (circles, parabolas, ellipses, hyperbolas) is the equation of a chord when its midpoint is known. This formula allows us to directly find the equation of the line segment without needing to first determine the endpoints of the chord.

Let the general equation of a conic section be represented by $S = 0$. For an ellipse, this is typically $S \equiv \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 = 0$.

If $P(x_1, y_1)$ is the midpoint of a chord of this conic section, then the equation of the chord is given by the formula: $T = S_1$ Let's break down what $T$ and $S_1$ mean:

• $T$ (Tangent-like expression): This is an expression derived from the general tangent equation. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $T$ is given by: $T \equiv \frac{x x_1}{a^2} + \frac{y y_1}{b^2} - 1$ It's important to note that this is not the equation of a tangent unless the point $(x_1, y_1)$ lies on the ellipse. When $(x_1, y_1)$ is the midpoint of a chord, it's typically inside the ellipse.

• $S_1$ (Value of the conic equation at the point): This is the numerical value obtained by substituting the coordinates of the given midpoint $(x_1, y_1)$ into the original ellipse equation $S$: $S_1 \equiv \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$ The value of $S_1$ tells us about the position of the point $(x_1, y_1)$ relative to the ellipse:

  • If $S_1 < 0$, the point is inside the ellipse (which must be true for a midpoint of a real chord).
  • If $S_1 = 0$, the point is on the ellipse (in which case the "chord" becomes a tangent, and $T=S_1$ reduces to $T=0$, the tangent equation).
  • If $S_1 > 0$, the point is outside the ellipse (it cannot be the midpoint of a real chord).

This formula is incredibly useful because it bypasses the need to find the endpoints of the chord first, which would involve solving a system of equations that is often more complex.

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Step-by-Step Solution

Step 1: Identify the Ellipse Equation and Midpoint Coordinates

First, we extract the given information from the problem: The equation of the ellipse is: $\frac{x^2}{4} + \frac{y^2}{2} = 1$ We compare this to the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to identify its parameters:

• $a^2 = 4$
• $b^2 = 2$

The coordinates of the midpoint of the chord are given as $P(x_1, y_1) = \left(1, \frac{1}{2}\right)$. So, $x_1 = 1$ and $y_1 = \frac{1}{2}$.

Step 2: Calculate the Components $T$ and $S_1$

Now, we will use the identified values of $a^2$, $b^2$, $x_1$, and $y_1$ to compute the expressions for $T$ and $S_1$.

2.1 Calculate $T$: Using the formula $T \equiv \frac{x x_1}{a^2} + \frac{y y_1}{b^2} - 1$: $T = \frac{x \cdot 1}{4} + \frac{y \cdot (1/2)}{2} - 1$ Simplify the expression: $T = \frac{x}{4} + \frac{y}{4} - 1$ This expression represents the left-hand side of the chord's equation.

2.2 Calculate $S_1$: Using the formula $S_1 \equiv \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} - 1$: $S_1 = \frac{(1)^2}{4} + \frac{(1/2)^2}{2} - 1$ Perform the squaring and division operations: $S_1 = \frac{1}{4} + \frac{1/4}{2} - 1$ $S_1 = \frac{1}{4} + \frac{1}{8} - 1$ To combine these fractions, find a common denominator, which is 8: $S_1 = \frac{2}{8} + \frac{1}{8} - \frac{8}{8}$ $S_1 = \frac{2 + 1 - 8}{8} = \frac{-5}{8}$ As anticipated, $S_1$ is negative, which confirms that the given midpoint $\left(1, \frac{1}{2}\right)$ lies inside the ellipse. This is a good sanity check.

Step 3: Form the Equation of the Chord using $T=S_1$

Now, we equate the calculated expressions for $T$ and $S_1$: $\frac{x}{4} + \frac{y}{4} - 1 = -\frac{5}{8}$ To simplify and eliminate fractions, multiply the entire equation by the least common multiple of the denominators (4 and 8), which is 8: $8 \left(\frac{x}{4} + \frac{y}{4} - 1\right) = 8 \left(-\frac{5}{8}\right)$ $2x + 2y - 8 = -5$ Rearrange the terms to express the equation in a standard linear form: $2x + 2y = 8 - 5$ $2x + 2y = 3$ This is the equation of the chord. We can also write it as $y = \frac{3-2x}{2} = \frac{3}{2} - x$. This form will be useful in the next step.

Step 4: Find the Intersection Points (Endpoints of the Chord)

To determine the length of the chord, we need the coordinates of its two endpoints. These points are the intersections of the chord (the line segment) with the ellipse. We find them by solving the equation of the chord and the equation of the ellipse simultaneously.

From the chord equation, we have $y = \frac{3}{2} - x$. Substitute this expression for $y$ into the ellipse equation $\frac{x^2}{4} + \frac{y^2}{2} = 1$: $\frac{x^2}{4} + \frac{\left(\frac{3}{2} - x\right)^2}{2} = 1$ To clear the denominators, multiply the entire equation by 4: $4 \left(\frac{x^2}{4}\right) + 4 \left(\frac{\left(\frac{3}{2} - x\right)^2}{2}\right) = 4 \cdot 1$ $x^2 + 2\left(\frac{3}{2} - x\right)^2 = 4$ Now, expand the squared term using the identity $(A-B)^2 = A^2 - 2AB + B^2$: $x^2 + 2\left(\left(\frac{3}{2}\right)^2 - 2 \cdot \frac{3}{2} \cdot x + x^2\right) = 4$ $x^2 + 2\left(\frac{9}{4} - 3x + x^2\right) = 4$ Distribute the 2: $x^2 + \frac{9}{2} - 6x + 2x^2 = 4$ Combine the like terms (terms with $x^2$, terms with $x$, and constant terms): $(x^2 + 2x^2) - 6x + \frac{9}{2} - 4 = 0$ $3x^2 - 6x + \left(\frac{9}{2} - \frac{8}{2}\right) = 0$ $3x^2 - 6x + \frac{1}{2} = 0$ To get rid of the fraction, multiply the entire quadratic equation by 2: $6x^2 - 12x + 1 = 0$ This quadratic equation's roots, let's call them $x_A$ and $x_B$, are the x-coordinates of the two endpoints of the chord.

Step 5: Apply Vieta's Formulas

Instead of directly solving the quadratic equation using the quadratic formula (which would give us the exact $x_A$ and $x_B$ values), we can use Vieta's formulas. This is often more efficient when we need expressions involving sums or products of roots, or differences of roots, rather than the roots themselves.

For a quadratic equation $Ax^2 + Bx + C = 0$, with roots $x_A$ and $x_B$:

• Sum of roots: $x_A + x_B = -\frac{B}{A}$
• Product of roots: $x_A x_B = \frac{C}{A}$

For our equation $6x^2 - 12x + 1 = 0$:

• $A=6$, $B=-12$, $C=1$
• Sum of x-coordinates: $x_A + x_B = -\frac{-12}{6} = 2$
• Product of x-coordinates: $x_A x_B = \frac{1}{6}$

Step 6: Calculate the Length of the Chord

Let the endpoints of the chord be $A(x_A, y_A)$ and $B(x_B, y_B)$. The length of the chord $L$ is given by the distance formula: $L = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}$ We know that both points $A$ and $B$ lie on the chord $y = \frac{3}{2} - x$. This means the slope of the chord is $m = -1$. For any two points on a line with slope $m$, the relationship between their coordinate differences is $y_B - y_A = m(x_B - x_A)$. In our case, $m = -1$, so: $y_B - y_A = -1(x_B - x_A)$ $y_B - y_A = -(x_B - x_A)$ Squaring both sides gives: $(y_B - y_A)^2 = (-(x_B - x_A))^2 = (x_B - x_A)^2$ Substitute this back into the length formula: $L = \sqrt{(x_B - x_A)^2 + (x_B - x_A)^2}$ $L = \sqrt{2(x_B - x_A)^2}$ $L = \sqrt{2} |x_B - x_A|$ This is a general result for a line with slope $m=-1$. For a general slope $m$, the length would be $L = \sqrt{1+m^2} |x_B - x_A|$.

Now, we need to find $|x_B - x_A|$. We can use the algebraic identity: $(x_B - x_A)^2 = (x_A + x_B)^2 - 4x_A x_B$ So, $|x_B - x_A| = \sqrt{(x_A + x_B)^2 - 4x_A x_B}$. Substitute the values obtained from Vieta's formulas: $x_A + x_B = 2$ and $x_A x_B = \frac{1}{6}$: $|x_B - x_A| = \sqrt{(2)^2 - 4\left(\frac{1}{6}\right)}$ $|x_B - x_A| = \sqrt{4 - \frac{4}{6}}$ $|x_B - x_A| = \sqrt{4 - \frac{2}{3}}$ To subtract, find a common denominator: $|x_B - x_A| = \sqrt{\frac{12}{3} - \frac{2}{3}}$ $|x_B - x_A| = \sqrt{\frac{10}{3}}$ Now, substitute this back into the chord length formula $L = \sqrt{2} |x_B - x_A|$: $L = \sqrt{2} \cdot \sqrt{\frac{10}{3}}$ $L = \sqrt{2 \cdot \frac{10}{3}}$ $L = \sqrt{\frac{20}{3}}$ To simplify the radical, we can write $20 = 4 \times 5$: $L = \sqrt{\frac{4 \cdot 5}{3}} = \frac{\sqrt{4} \cdot \sqrt{5}}{\sqrt{3}} = \frac{2\sqrt{5}}{\sqrt{3}}$ To rationalize