The locus of mid-points of the line segments joining (-3, -5) and the points on the ellipse x²/4 + y²/9 = 1 is :

Solution

Maths sir · 5 steps

Take a general point on the ellipse as (2 cos θ, 3 sin θ), find the midpoint, then remove θ.

1. General point. Here a = 2 and b = 3, so any point on the ellipse is (2 cos θ, 3 sin θ). Let the midpoint of this point and (-3, -5) be (h, k).
2. Midpoint. h = (2 cos θ - 3)/2 and k = (3 sin θ - 5)/2. Solving for the trig terms gives cos θ = (2h+3)/2 and sin θ = (2k+5)/3.
3. Remove theta. Use cos² θ + sin² θ = 1 and multiply through by 36:

   9(2h+3)² + 4(2k+5)² = 36

4. Expand. 36h² + 108h + 81 + 16k² + 80k + 100 - 36 = 0, which is 36h² + 16k² + 108h + 80k + 145 = 0.
5. So. Replacing (h, k) by (x, y), the locus is 36x² + 16y² + 108x + 80y + 145 = 0, which is option (c).

The idea to keep. For a locus built from points on a conic, write the point in parametric form, express the parameter's sine and cosine in terms of h and k, then use sin² θ + cos² θ = 1.

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