The locus of the mid-points of the lines joining the extremities of two semi-conjugate diameters of an ellipse is ..........

Answer

To find the locus of the midpoints of the lines joining the extremities of two semi-conjugate diameters of an ellipse, we can follow these steps:

### Step 1: Understand the Ellipse and Semi-Conjugate Diameters
The standard equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. Semi-conjugate diameters are diameters that are perpendicular to each other.

### Step 2: Parametric Representation of Points
The points on the ellipse can be represented parametrically as:
- For point \(P\): \(P(a \cos \theta, b \sin \theta)\)
- For point \(Q\): \(Q(-b \sin \theta, a \cos \theta)\)

### Step 3: Find the Midpoint of the Line Joining Points \(P\) and \(Q\)
The midpoint \(M\) of the line segment joining points \(P\) and \(Q\) is given by:
\[ M\left(H, K\right) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Substituting the coordinates of points \(P\) and \(Q\):
\[ H = \frac{a \cos \theta - b \sin \theta}{2} \]
\[ K = \frac{b \sin \theta + a \cos \theta}{2} \]

### Step 4: Express \(H\) and \(K\) in Terms of \(\theta\)
From the expressions for \(H\) and \(K\), we can manipulate them:
1. Multiply the equation for \(H\) by 2:
\[ 2H = a \cos \theta - b \sin \theta \quad \text{(1)} \]
2. Multiply the equation for \(K\) by 2:
\[ 2K = b \sin \theta + a \cos \theta \quad \text{(2)} \]

### Step 5: Square and Add the Equations
Now, we will square both equations and add them:
- From equation (1):
\[ (2H)^2 = (a \cos \theta - b \sin \theta)^2 \]
- From equation (2):
\[ (2K)^2 = (b \sin \theta + a \cos \theta)^2 \]

### Step 6: Expand and Simplify
Expanding both squared terms:
1. For \(H\):
\[ 4H^2 = a^2 \cos^2 \theta - 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta \]
2. For \(K\):
\[ 4K^2 = b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta + a^2 \cos^2 \theta \]

Adding these two equations:
\[ 4H^2 + 4K^2 = (a^2 \cos^2 \theta + b^2 \sin^2 \theta) + (a^2 \cos^2 \theta + b^2 \sin^2 \theta) = a^2 + b^2 \]

### Step 7: Final Locus Equation
Thus, we have:
\[ 4H^2 + 4K^2 = a^2 + b^2 \]
Dividing through by \(a^2 + b^2\):
\[ \frac{4H^2}{a^2} + \frac{4K^2}{b^2} = 1 \]
This can be rewritten as:
\[ \frac{H^2}{\frac{a^2}{4}} + \frac{K^2}{\frac{b^2}{4}} = 1 \]

### Conclusion
The locus of the midpoints of the lines joining the extremities of two semi-conjugate diameters of an ellipse is an ellipse given by:
\[ \frac{x^2}{\frac{a^2}{4}} + \frac{y^2}{\frac{b^2}{4}} = 1 \]