CP and CD are conjugate semi-diameters of the ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1`, The locus of the mid-point of PD, is

Options

A. `(x^2)/(a^2)+(y^2)/(b^2)=(1)/(2)`
B. `(x^2)/(a^2)+(y^2)/(b^2)=1`
C. `(x^2)/(a^2)+(y^2)/(b^2)=(3)/(2)`
D. `(x^2)/(a^2)+(y^2)/(b^2)=2`

Answer

To find the locus of the midpoint of the line segment PD where CP and CD are conjugate semi-diameters of the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step 1: Identify Points P and D
Given that CP and CD are conjugate semi-diameters, we can express the coordinates of points P and D in terms of a parameter \( \theta \):
- Point P: \( P(a \cos \theta, b \sin \theta) \)
- Point D: Since D is conjugate to P, its coordinates can be expressed as:
\[ D(-a \sin \theta, b \cos \theta) \]

### Step 2: Find the Midpoint M of PD
The midpoint M of the line segment PD can be calculated using the midpoint formula:
\[ M(h, k) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of P and D:
\[ h = \frac{a \cos \theta - a \sin \theta}{2} = \frac{a(\cos \theta - \sin \theta)}{2} \]
\[ k = \frac{b \sin \theta + b \cos \theta}{2} = \frac{b(\sin \theta + \cos \theta)}{2} \]

### Step 3: Express \( h \) and \( k \) in Terms of \( \theta \)
From the expressions for \( h \) and \( k \), we can write:
\[ \frac{2h}{a} = \cos \theta - \sin \theta \]
\[ \frac{2k}{b} = \sin \theta + \cos \theta \]

### Step 4: Square and Add the Equations
Now we square both equations and add them:
\[ \left(\frac{2h}{a}\right)^2 + \left(\frac{2k}{b}\right)^2 = (\cos \theta - \sin \theta)^2 + (\sin \theta + \cos \theta)^2 \]
Calculating the right-hand side:
\[ (\cos \theta - \sin \theta)^2 = \cos^2 \theta - 2\cos \theta \sin \theta + \sin^2 \theta \]
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta \]
Adding these two results:
\[ \cos^2 \theta + \sin^2 \theta - 2\cos \theta \sin \theta + \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 2 \]
Thus, we have:
\[ \left(\frac{2h}{a}\right)^2 + \left(\frac{2k}{b}\right)^2 = 2 \]

### Step 5: Simplify the Equation
This can be rewritten as:
\[ \frac{h^2}{\left(\frac{a}{\sqrt{2}}\right)^2} + \frac{k^2}{\left(\frac{b}{\sqrt{2}}\right)^2} = 1 \]
This represents an ellipse with semi-major axis \( \frac{a}{\sqrt{2}} \) and semi-minor axis \( \frac{b}{\sqrt{2}} \).

### Final Equation
Replacing \( h \) and \( k \) with \( x \) and \( y \), we get the locus of the midpoint PD:
\[ \frac{x^2}{\frac{a^2}{2}} + \frac{y^2}{\frac{b^2}{2}} = 1 \]
or
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{2} \]

### Conclusion
The locus of the midpoint of PD is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{1}{2} \]