Question

The locus of mid-points of the line segments joining \( (-3, -5) \) and the points on the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) is :

• A. \( 36x^2 + 16y^2 + 90x + 56y + 145 = 0 \)
• B. \( 9x^2 + 4y^2 + 18x + 8y + 145 = 0 \)
• C. \( 36x^2 + 16y^2 + 72x + 32y + 145 = 0 \)
• D. \( 36x^2 + 16y^2 + 108x + 80y + 145 = 0 \)

Hint:

The locus of midpoints of segments from a fixed point to a conic is always another conic of the same type.
If the original conic is \( f(x, y) = 0 \), the locus of midpoints with respect to \( (x_0, y_0) \) is \( f(2x-x_0, 2y-y_0) = 0 \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Let \( P(x_1, y_1) \) be any point on the ellipse.
Let \( Q(h, k) \) be the midpoint of the segment joining \( A(-3, -5) \) and \( P(x_1, y_1) \).
We express \( x_1, y_1 \) in terms of \( h, k \) and substitute into the ellipse equation.
Step 2: Key Formula or Approach:
Midpoint formula: \( h = \frac{x_1 - 3}{2}, k = \frac{y_1 - 5}{2} \).
Step 3: Detailed Explanation:
From midpoint equations:
\( x_1 = 2h + 3 \)
\( y_1 = 2k + 5 \)
Since \( (x_1, y_1) \) lies on the ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \):
\[ \frac{(2h + 3)^2}{4} + \frac{(2k + 5)^2}{9} = 1 \]
Multiplying the entire equation by 36:
\[ 9(2h + 3)^2 + 4(2k + 5)^2 = 36 \]
\[ 9(4h^2 + 12h + 9) + 4(4k^2 + 20k + 25) = 36 \]
\[ 36h^2 + 108h + 81 + 16k^2 + 80k + 100 = 36 \]
\[ 36h^2 + 16k^2 + 108h + 80k + 181 - 36 = 0 \]
\[ 36h^2 + 16k^2 + 108h + 80k + 145 = 0 \]
Replacing \( (h, k) \) with \( (x, y) \):
\[ 36x^2 + 16y^2 + 108x + 80y + 145 = 0 \]
Step 4: Final Answer:
The locus is \( 36x^2 + 16y^2 + 108x + 80y + 145 = 0 \).