Find the locus of the middle points of chords of an ellipse whose length is constant ( = 2c).

Answer

To find the locus of the midpoints of chords of an ellipse whose length is constant (2c), we can follow these steps:

### Step 1: Understand the Ellipse Equation
The standard equation of an ellipse centered at the origin 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.

### Step 2: Define the Chord Length
Let the length of the chord be \(2c\). The midpoint of the chord can be denoted as \(P(\alpha, \beta)\).

### Step 3: Set Up the Coordinates of the Endpoints of the Chord
Let the endpoints of the chord be \(A(x_1, y_1)\) and \(B(x_2, y_2)\). The midpoint \(P\) is given by:
\[ P\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = (\alpha, \beta) \]
From this, we can express the coordinates of \(A\) and \(B\):
\[ x_1 = \alpha - u, \quad y_1 = \beta - v \]
\[ x_2 = \alpha + u, \quad y_2 = \beta + v \]
where \(u\) and \(v\) are the horizontal and vertical distances from the midpoint to the endpoints.

### Step 4: Use the Distance Formula
The length of the chord \(AB\) is given by the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 2c \]
Substituting for \(x_1\), \(x_2\), \(y_1\), and \(y_2\):
\[ \sqrt{((\alpha + u) - (\alpha - u))^2 + ((\beta + v) - (\beta - v))^2} = 2c \]
This simplifies to:
\[ \sqrt{(2u)^2 + (2v)^2} = 2c \]
Thus:
\[ \sqrt{4u^2 + 4v^2} = 2c \]
Squaring both sides gives:
\[ 4u^2 + 4v^2 = 4c^2 \]
Dividing by 4:
\[ u^2 + v^2 = c^2 \]

### Step 5: Relate \(u\) and \(v\) to the Ellipse
Since points \(A\) and \(B\) lie on the ellipse, we can substitute \(x_1\) and \(y_1\) into the ellipse equation:
\[ \frac{(\alpha - u)^2}{a^2} + \frac{(\beta - v)^2}{b^2} = 1 \]
And for point \(B\):
\[ \frac{(\alpha + u)^2}{a^2} + \frac{(\beta + v)^2}{b^2} = 1 \]

### Step 6: Set Up the Locus Equation
By adding these two equations, we can eliminate \(u\) and \(v\):
\[ \frac{(\alpha - u)^2 + (\alpha + u)^2}{a^2} + \frac{(\beta - v)^2 + (\beta + v)^2}{b^2} = 2 \]
This simplifies to:
\[ \frac{2\alpha^2 + 2u^2}{a^2} + \frac{2\beta^2 + 2v^2}{b^2} = 2 \]
Dividing by 2:
\[ \frac{\alpha^2 + u^2}{a^2} + \frac{\beta^2 + v^2}{b^2} = 1 \]
Substituting \(u^2 + v^2 = c^2\):
\[ \frac{\alpha^2 + c^2}{a^2} + \frac{\beta^2 + c^2}{b^2} = 1 \]

### Step 7: Final Locus Equation
Rearranging gives us the locus of the midpoint \(P(\alpha, \beta)\):
\[ \frac{\alpha^2}{a^2} + \frac{\beta^2}{b^2} = 1 - \frac{c^2}{a^2} - \frac{c^2}{b^2} \]
This represents an ellipse.