Question

The locus of the mid-point of the line segment joining the point (4, 3) and the points on the ellipse x² + 2y² = 4 is an ellipse with eccentricity:

• A. \(\frac{\sqrt3}{2}\)
• B. \(\frac{1}{2\sqrt2}\)
• C. \(\frac{1}{\sqrt2}\)
• D. \(\frac{1}{2}\)

Answer 1

The Correct Option is C

To determine the locus of the midpoint of the line segment joining the fixed point \((4, 3)\) with any point \((x_1, y_1)\) on the ellipse \(x^2 + 2y^2 = 4\), follow these steps:

1. Let \((x_m, y_m)\) be the midpoint of the line segment joining \((4, 3)\) and \((x_1, y_1)\). The coordinates of the midpoint are given by:
   x_m = \frac{x_1 + 4}{2}
   y_m = \frac{y_1 + 3}{2}
2. Solving the above expressions for \(x_1\) and \(y_1\), we get:
   x_1 = 2x_m - 4
   y_1 = 2y_m - 3
3. Since \((x_1, y_1)\) lies on the ellipse \(x^2 + 2y^2 = 4\), substitute \(x_1\) and \(y_1\) into the ellipse equation:
   (2x_m - 4)^2 + 2(2y_m - 3)^2 = 4
4. Expanding and simplifying the equation:
   4x_m^2 - 16x_m + 16 + 8y_m^2 - 24y_m + 18 = 4
   4x_m^2 + 8y_m^2 - 16x_m - 24y_m + 30 = 0
5. Divide the entire equation by 2 to simplify:
   2x_m^2 + 4y_m^2 - 8x_m - 12y_m + 15 = 0
6. Recognize that this is the equation of an ellipse. To find its eccentricity, put it in the standard form by completing the square:
   2(x_m^2 - 4x_m) + 4(y_m^2 - 3y_m) = -15
   Completing the square:
   2((x_m - 2)^2 - 4) + 4((y_m - \frac{3}{2})^2 - \frac{9}{4}) = -15
   2(x_m - 2)^2 + 4(y_m - \frac{3}{2})^2 = 1
7. Divide by 2 to convert it to standard form:
   \frac{(x_m - 2)^2}{\frac{1}{2}} + \frac{(y_m - \frac{3}{2})^2}{\frac{1}{4}} = 1
8. Identify the parameters of the ellipse: semi-major axis \(a = \frac{1}{\sqrt{2}}\), semi-minor axis \(b = \frac{1}{2}\).
9. Calculate the eccentricity \(e\) using the formula:
   e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/4}{1/2}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

Therefore, the eccentricity of the ellipse is \(\frac{1}{\sqrt{2}}\).