1. Understanding the Problem and Key Concepts

This problem asks us to find the locus of the mid-points of line segments. A locus is the set of all points that satisfy a given geometric condition. Here, the condition is that each point on the locus is the midpoint of a line segment connecting a fixed point and a variable point on an ellipse.

To solve this, we will use the following fundamental concepts:

• Midpoint Formula: For two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the coordinates of their midpoint $M(x_M, y_M)$ are given by: $x_M = \frac{x_1 + x_2}{2}, \quad y_M = \frac{y_1 + y_2}{2}$
• Parametric Form of an Ellipse: For an ellipse given by the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, any point $(x_E, y_E)$ on the ellipse can be conveniently represented using a single parameter, $\theta$ (the eccentric angle), as: $(x_E, y_E) = (a\cos\theta, b\sin\theta)$ This representation is extremely useful in locus problems involving points on an ellipse because it allows us to express both coordinates using a single variable, which can then be eliminated to find the locus equation.
• Elimination of Parameter: Once we express the coordinates of the locus point in terms of $\theta$, we will use trigonometric identities (specifically $\cos^2\theta + \sin^2\theta = 1$) to eliminate $\theta$ and obtain an equation solely in terms of the locus coordinates.

2. Step-by-Step Derivation of the Locus

Step 1: Identify the fixed point and the equation of the ellipse. We are given a fixed point, let's call it $P_1$: $P_1 = (-3, -5)$ The equation of the ellipse is: $\frac{x^2}{4} + \frac{y^2}{9} = 1$ From this equation, we can identify $a^2 = 4$ and $b^2 = 9$. This means $a = 2$ and $b = 3$.

Step 2: Represent a general point on the ellipse using its parametric form. Let $P_2$ be any point on the ellipse. Using the parametric form $(a\cos\theta, b\sin\theta)$, we can write $P_2$ as: $P_2 = (2\cos\theta, 3\sin\theta)$ Why this step? By using the parametric form, we represent all possible points on the ellipse using a single variable $\theta$. This simplifies the subsequent algebraic manipulations when finding the midpoint.

Step 3: Define the midpoint and apply the midpoint formula. Let $M(h, k)$ be the midpoint of the line segment joining $P_1(-3, -5)$ and $P_2(2\cos\theta, 3\sin\theta)$. Using the midpoint formula: $h = \frac{-3 + 2\cos\theta}{2}$ $k = \frac{-5 + 3\sin\theta}{2}$ Why this step? We are looking for the locus of these midpoints. By denoting a general midpoint as $(h, k)$, we establish the relationship between its coordinates and the parameter $\theta$ (which defines the points on the ellipse).

Step 4: Express $\cos\theta$ and $\sin\theta$ in terms of $h$ and $k$. From the midpoint equations, we need to isolate $\cos\theta$ and $\sin\theta$ so we can use the trigonometric identity. From the equation for $h$: $2h = -3 + 2\cos\theta$ $2\cos\theta = 2h + 3$ $\cos\theta = \frac{2h + 3}{2} \quad \text{--- (1)}$ From the equation for $k$: $2k = -5 + 3\sin\theta$ $3\sin\theta = 2k + 5$ $\sin\theta = \frac{2k + 5}{3} \quad \text{--- (2)}$ Why this step? Our goal is to find an equation relating $h$ and $k$ that does not depend on $\theta$. By expressing $\cos\theta$ and $\sin\theta$ in terms of $h$ and $k$, we prepare for the elimination of $\theta$.

Step 5: Eliminate the parameter $\theta$ using a trigonometric identity. We know the fundamental trigonometric identity: $\cos^2\theta + \sin^2\theta = 1$ Substitute the expressions for $\cos\theta$ and $\sin\theta$ from equations (1) and (2) into this identity: $\left(\frac{2h + 3}{2}\right)^2 + \left(\frac{2k + 5}{3}\right)^2 = 1$ Why this step? This is the crucial step to eliminate the parameter $\theta$. The resulting equation will directly describe the relationship between $h$ and $k$, which is the locus we are seeking.

Step 6: Simplify the equation to get the locus. Now, we expand and simplify the equation: $\frac{(2h + 3)^2}{4} + \frac{(2k + 5)^2}{9} = 1$ To remove the denominators, find the least common multiple (LCM) of 4 and 9, which is 36. Multiply the entire equation by 36: $9(2h + 3)^2 + 4(2k + 5)^2 = 36$ Expand the squared terms: $9(4h^2 + 12h + 9) + 4(4k^2 + 20k + 25) = 36$ Distribute the coefficients: $36h^2 + 108h + 81 + 16k^2 + 80k + 100 = 36$ Rearrange the terms to match the standard form of an ellipse and move the constant term to the left side: $36h^2 + 16k^2 + 108h + 80k + 81 + 100 - 36 = 0$ $36h^2 + 16k^2 + 108h + 80k + 145 = 0$

Step 7: Replace $(h, k)$ with $(x, y)$ to represent the locus. The final step is to replace $h$ with $x$ and $k$ with $y$ to express the locus in standard coordinate notation: $36x^2 + 16y^2 + 108x + 80y + 145 = 0$

3. Comparing with Options

Let's check the options provided: (A) $9{x^2} + 4{y^2} + 18x + 8y + 145 = 0$ (B) $36{x^2} + 16{y^2} + 90x + 56y + 145 = 0$ (C) $36{x^2} + 16{y^2} + 108x + 80y + 145 = 0$ (D) $36{x^2} + 16{y^2} + 72x + 32y + 145 = 0$

Our derived equation $36x^2 + 16y^2 + 108x + 80y + 145 = 0$ matches option (C).

Wait! Let's recheck the question and options carefully. The correct answer is given as (A). Let's re-evaluate my steps.

My derived equation: $36x^2 + 16y^2 + 108x + 80y + 145 = 0$. Option (A): $9x^2 + 4y^2 + 18x + 8y + 145 = 0$.

There is a discrepancy in the coefficients of $x^2, y^2, x, y$. The constant term $145$ matches. Let's divide my equation by 4: $\frac{36x^2}{4} + \frac{16y^2}{4} + \frac{108x}{4} + \frac{80y}{4} + \frac{145}{4} = 0$ $9x^2 + 4y^2 + 27x + 20y + \frac{145}{4} = 0$ This doesn't match option (A).

Let's re-check the calculation: $9(2h + 3)^2 + 4(2k + 5)^2 = 36$ $9(4h^2 + 12h + 9) + 4(4k^2 + 20k + 25) = 36$ $36h^2 + 108h + 81 + 16k^2 + 80k + 100 = 36$ $36h^2 + 16k^2 + 108h + 80k + 181 = 36$ $36h^2 + 16k^2 + 108h + 80k + 181 - 36 = 0$ $36h^2 + 16k^2 + 108h + 80k + 145 = 0$

The calculation is correct. This leads to option (C). The provided "Correct Answer: A" seems to be incorrect based on my derivation. Let's consider if the question intended the ellipse to be $\frac{x^2}{9} + \frac{y^2}{4} = 1$ instead. If $a=3, b=2$, then $P_2 = (3\cos\theta, 2\sin\theta)$. $h = \frac{-3 + 3\cos\theta}{2} \implies \cos\theta = \frac{2h+3}{3}$ $k = \frac{-5 + 2\sin\theta}{2} \implies \sin\theta = \frac{2k+5}{2}$ $\left(\frac{2h+3}{3}\right)^2 + \left(\frac{2k+5}{2}\right)^2 = 1$ $\frac{(2h+3)^2}{9} + \frac{(2k+5)^2}{4} = 1$ Multiply by 36: $4(2h+3)^2 + 9(2k+5)^2 = 36$ $4(4h^2+12h+9) + 9(4k^2+20k+25) = 36$ $16h^2+48h+36 + 36k^2+180k+225 = 36$ $16h^2+36k^2+48h+180k+261 = 36$ $16h^2+36k^2+48h+180k+225 = 0$ This does not match option (A) either.

Let's re-examine option (A): $9x^2 + 4y^2 + 18x + 8y + 145 = 0$. If this is the answer, then it must be derived from: $\frac{(x+1)^2}{4} + \frac{(y+1)^2}{9} = 1$ (This would be an ellipse centered at $(-1,-1)$ with $a=2, b=3$) or $\frac{(x+1)^2}{9} + \frac{(y+1)^2}{4} = 1$ (This would be an ellipse centered at $(-1,-1)$ with $a=3, b=2$) Let's check the first case: $\frac{(x+1)^2}{4} + \frac{(y+1)^2}{9} = 1$ $9(x+1)^2 + 4(y+1)^2 = 36$ $9(x^2+2x+1) + 4(y^2+2y+1) = 36$ $9x^2+18x+9 + 4y^2+8y+4 = 36$ $9x^2+4y^2+18x+8y+13 = 36$ $9x^2+4y^2+18x+8y-23 = 0$ This is not option (A). The constant term is vastly different.

Let's assume there is a typo in the question or options or the provided correct answer. My derivation is sound for the given question. The process for finding the locus of midpoints is:

1. Define the fixed point $P_1(x_1, y_1)$.
2. Define a general point on the curve $P_2(x_2(\theta), y_2(\theta))$.
3. Let the midpoint be $M(h, k)$.
4. Use the midpoint formula: $h = \frac{x_1 + x_2(\theta)}{2}$, $k = \frac{y_1 + y_2(\theta)}{2}$.
5. Solve for $x_2(\theta)$ and $y_2(\theta)$ in terms of $h$ and $k$.
6. Substitute these into the equation of the original curve (or a relation like $\cos^2\theta + \sin^2\theta = 1$).
7. Replace $h$ with $x$ and $k$ with $y$.

Let's follow this logic strictly. Given fixed point $P_1(-3, -5)$. Let $P_2(X, Y)$ be a point on the ellipse $\frac{X^2}{4} + \frac{Y^2}{9} = 1$. Let $M(x, y)$ be the midpoint of $P_1P