The line `y=x+rho` (p is parameter) cuts the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at P and Q, then prove that mid point of PQ lies on `a^(2)y=-b^(2)x`.

Answer

To solve the problem step by step, we need to find the midpoint of the points P and Q where the line \( y = x + p \) intersects the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We will then show that this midpoint lies on the line \( a^2y = -b^2x \).

### Step 1: Substitute the line equation into the ellipse equation

Given the line \( y = x + p \), we can substitute this into the ellipse equation:

\[ \frac{x^2}{a^2} + \frac{(x + p)^2}{b^2} = 1 \]

### Step 2: Expand and simplify the equation

Expanding \( (x + p)^2 \):

\[ \frac{x^2}{a^2} + \frac{x^2 + 2px + p^2}{b^2} = 1 \]

Combining terms:

\[ \frac{x^2}{a^2} + \frac{x^2}{b^2} + \frac{2px}{b^2} + \frac{p^2}{b^2} = 1 \]

### Step 3: Multiply through by \( a^2b^2 \) to eliminate denominators

\[ b^2x^2 + a^2x^2 + 2apx + a^2p^2 = a^2b^2 \]

Combining the \( x^2 \) terms:

\[ (b^2 + a^2)x^2 + 2apx + (a^2p^2 - a^2b^2) = 0 \]

### Step 4: Use the quadratic formula to find the x-coordinates of P and Q

This is a quadratic equation in \( x \):

\[ Ax^2 + Bx + C = 0 \]

Where:
- \( A = b^2 + a^2 \)
- \( B = 2ap \)
- \( C = a^2p^2 - a^2b^2 \)

The roots \( x_1 \) and \( x_2 \) (the x-coordinates of points P and Q) can be found using the quadratic formula:

\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]

### Step 5: Find the midpoint of P and Q

The x-coordinate of the midpoint \( M \) is:

\[ x_m = \frac{x_1 + x_2}{2} = \frac{-B}{2A} = \frac{-2ap}{2(b^2 + a^2)} = \frac{-ap}{b^2 + a^2} \]

The y-coordinates of points P and Q can be found by substituting \( x_1 \) and \( x_2 \) back into the line equation:

\[ y_1 = x_1 + p, \quad y_2 = x_2 + p \]

Thus, the y-coordinate of the midpoint \( M \) is:

\[ y_m = \frac{y_1 + y_2}{2} = \frac{(x_1 + p) + (x_2 + p)}{2} = \frac{(x_1 + x_2) + 2p}{2} = \frac{-B + 2p}{2} = \frac{-2ap + 2p}{2(b^2 + a^2)} = \frac{(2p(1 - a))}{2(b^2 + a^2)} = \frac{p(1 - a)}{b^2 + a^2} \]

### Step 6: Show that the midpoint lies on the line \( a^2y = -b^2x \)

Substituting \( x_m \) and \( y_m \) into the line equation \( a^2y = -b^2x \):

\[ a^2 \left(\frac{p(1 - a)}{b^2 + a^2}\right) = -b^2 \left(\frac{-ap}{b^2 + a^2}\right) \]

This simplifies to:

\[ \frac{a^2p(1 - a)}{b^2 + a^2} = \frac{ab^2p}{b^2 + a^2} \]

Since both sides are equal, we conclude that the midpoint \( M \) lies on the line \( a^2y = -b^2x \).