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 :

Solution

Maths sir · 4 steps

Write the ellipse point in terms of the midpoint, substitute, and read the new ellipse's shape.

1. Name the midpoint. Let (h, k) be the midpoint of (4, 3) and a point (x₁, y₁) on the ellipse. Then h = (x₁ + 4)/2 and k = (y₁ + 3)/2, so x₁ = 2h - 4 and y₁ = 2k - 3.
2. Substitute. Since (x₁, y₁) lies on x² + 2y² = 4, we get (2h-4)² + 2(2k-3)² = 4. Take out the factor 4 from each square and divide by 4.

   (h-2)²/1 + (k-3/2)²/(1/2) = 1

3. Find eccentricity. Here a² = 1 and b² = 1/2, so e² = 1 - b²/a² = 1 - 1/2 = 1/2.
4. So. e = 1/√2, which is option (c). It matches the original ellipse x²/4 + y²/2 = 1, because the midpoint locus is that ellipse shrunk by a factor of 1/2, and shrinking does not change the shape.

The idea to keep. Taking midpoints with a fixed point scales a curve by 1/2 about that point. Scaling changes size, not shape, so the eccentricity stays the same.

Answer from the NTA final answer key. Solution by PrepWiser.