Substitute x_1 = 1 and y_1 = (1)/(2) into the expressions for T and S_1. The equation for T is (xx_1)/(4) + (yy_1)/(2) - 1, and for S_1 it is (x_1²)/(4) + (y_1²)/(2) - 1.

Simplify both sides of the equation. The left side becomes (x)/(4) + (y)/(4) - 1. The right side becomes (1)/(4) + (1)/(8) - 1 = (2+1-8)/(8) = -(5)/(8). So, (x)/(4) + (y)/(4) - 1 = -(5)/(8). Adding 1 to both sides gives (x)/(4) + (y)/(4) = 1 - (5)/(8) = (3)/(8). Multiplying by 8 gives 2x + 2y = 3. This is the equation of the chord.

Simplify both sides of the equation. The left side becomes (x)/(4) + (y)/(4) - 1. The right side becomes (1)/(4) + (1)/(8) - 1 = (2+1-8)/(8) = -(5)/(8). So, (x)/(4) + (y)/(4) - 1 = -(5)/(8). Adding 1 to both sides gives (x)/(4) + (y)/(4) = 1 - (5)/(8) = (3)/(8). Multiplying by 8 gives 2x + 2y = 3. This is the equation of the chord.

To find the length of the chord, we need its endpoints. We can find these by solving the equation of the chord (2x + 2y = 3 ⇒ y = (3-2x)/(2)) simultaneously with the equation of the ellipse (x²)/(4) + (y²)/(2) = 1. Substitute the expression for y from the chord equation into the ellipse equation.

To find the length of the chord, we need its endpoints. We can find these by solving the equation of the chord (2x + 2y = 3 ⇒ y = (3-2x)/(2)) simultaneously with the equation of the ellipse (x²)/(4) + (y²)/(2) = 1. Substitute the expression for y from the chord equation into the ellipse equation.

Simplify the equation: (x²)/(4) + (9 - 12x + 4x²)/(8) = 1. Multiply by 8 to clear denominators: 2x² + 9 - 12x + 4x² = 8. Rearrange into a quadratic equation: 6x² - 12x + 1 = 0. Let the roots be x_1 and x_2.

Simplify the equation: (x²)/(4) + (9 - 12x + 4x²)/(8) = 1. Multiply by 8 to clear denominators: 2x² + 9 - 12x + 4x² = 8. Rearrange into a quadratic equation: 6x² - 12x + 1 = 0. Let the roots be x_1 and x_2.

The length of the chord L is given by the distance formula. For the quadratic equation ax²+bx+c=0, (x_2-x_1)² = (x_1+x_2)² - 4x_1x_2 = ( -b/a)² - 4(c/a). For 6x² - 12x + 1 = 0, x_1+x_2 = 12/6 = 2 and x_1x_2 = 1/6. So (x_2-x_1)² = (2)² - 4(1/6) = 4 - 2/3 = 10/3. For 6y² - 6y + 1 = 0, y_1+y_2 = 6/6 = 1 and y_1y_2 = 1/6. So (y_2-y_1)² = (1)² - 4(1/6) = 1 - 2/3 = 1/3. Therefore, L = √((10)/(3) + (1)/(3)) = √((11)/(3)).

The length of the chord L is given by the distance formula. For the quadratic equation ax²+bx+c=0, (x_2-x_1)² = (x_1+x_2)² - 4x_1x_2 = ( -b/a)² - 4(c/a). For 6x² - 12x + 1 = 0, x_1+x_2 = 12/6 = 2 and x_1x_2 = 1/6. So (x_2-x_1)² = (2)² - 4(1/6) = 4 - 2/3 = 10/3. For 6y² - 6y + 1 = 0, y_1+y_2 = 6/6 = 1 and y_1y_2 = 1/6. So (y_2-y_1)² = (1)² - 4(1/6) = 1 - 2/3 = 1/3. Therefore, L = √((10)/(3) + (1)/(3)) = √((11)/(3)).

The length of the chord of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ , whose mid-point is $\left(1, \frac{1}{2}\right)$ , is:

Answer: The length of the chord is $\sqrt{\frac{11}{3}}$ .

Step-by-step solution

Step 1: Equation of the chord with a given midpoint

The equation of a chord of an ellipse with a given midpoint $(x_1, y_1)$ can be found using the formula $T = S_1$ . Here, $T$ represents the tangent equation at $(x_1, y_1)$ and $S_1$ represents the value of the ellipse equation when $(x, y)$ is replaced by $(x_1, y_1)$ . The given ellipse is $\frac{x^2}{4} + \frac{y^2}{2} = 1$ , and the midpoint is $\left(1, \frac{1}{2}\right)$ .

Step 2: Calculate T and S1

Substitute $x_1 = 1$ and $y_1 = \frac{1}{2}$ into the expressions for $T$ and $S_1$ . The equation for $T$ is $\frac{xx_1}{4} + \frac{yy_1}{2} - 1$ , and for $S_1$ it is $\frac{x_1^2}{4} + \frac{y_1^2}{2} - 1$ .

Step 3: Simplify the chord equation

Simplify both sides of the equation. The left side becomes $\frac{x}{4} + \frac{y}{4} - 1$ . The right side becomes $\frac{1}{4} + \frac{1}{8} - 1 = \frac{2+1-8}{8} = -\frac{5}{8}$ . So, $\frac{x}{4} + \frac{y}{4} - 1 = -\frac{5}{8}$ . Adding 1 to both sides gives $\frac{x}{4} + \frac{y}{4} = 1 - \frac{5}{8} = \frac{3}{8}$ . Multiplying by 8 gives $2x + 2y = 3$ . This is the equation of the chord.

Step 4: Find intersection points of the chord and ellipse

To find the length of the chord, we need its endpoints. We can find these by solving the equation of the chord ( $2x + 2y = 3 \implies y = \frac{3-2x}{2}$ ) simultaneously with the equation of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ . Substitute the expression for $y$ from the chord equation into the ellipse equation.

Step 5: Solve for x-coordinates of intersection points

Simplify the equation: $\frac{x^2}{4} + \frac{9 - 12x + 4x^2}{8} = 1$ . Multiply by 8 to clear denominators: $2x^2 + 9 - 12x + 4x^2 = 8$ . Rearrange into a quadratic equation: $6x^2 - 12x + 1 = 0$ . Let the roots be $x_1$ and $x_2$ .

Step 6: Solve for y-coordinates of intersection points

Similarly, we can express $x$ in terms of $y$ from the chord equation: $x = \frac{3-2y}{2}$ . Substitute this into the ellipse equation: $\frac{(\frac{3-2y}{2})^2}{4} + \frac{y^2}{2} = 1$ . This simplifies to $6y^2 - 6y + 1 = 0$ . Let the roots be $y_1$ and $y_2$ .

Step 7: Calculate the length of the chord

The length of the chord $L$ is given by the distance formula. For the quadratic equation $ax^2+bx+c=0$ , $(x_2-x_1)^2 = (x_1+x_2)^2 - 4x_1x_2 = ( -b/a )^2 - 4(c/a)$ . For $6x^2 - 12x + 1 = 0$ , $x_1+x_2 = 12/6 = 2$ and $x_1x_2 = 1/6$ . So $(x_2-x_1)^2 = (2)^2 - 4(1/6) = 4 - 2/3 = 10/3$ . For $6y^2 - 6y + 1 = 0$ , $y_1+y_2 = 6/6 = 1$ and $y_1y_2 = 1/6$ . So $(y_2-y_1)^2 = (1)^2 - 4(1/6) = 1 - 2/3 = 1/3$ . Therefore, $L = \sqrt{\frac{10}{3} + \frac{1}{3}} = \sqrt{\frac{11}{3}}$ .

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The length of the chord of the ellipse x24+y22=1\frac{x^2}{4} + \frac{y^2}{2} = 14x2​+2y2​=1, whose mid-point is (1,12)\left(1, \frac{1}{2}\right)(1,21​), is: