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, we need to prove that the midpoint of the points P and Q, where the line \( y = x + \rho \) intersects the ellipse given by the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \]

lies on the line defined by

\[ a^2 y = -b^2 x. \]

### Step-by-Step Solution

**Step 1: Substitute the line equation into the ellipse equation.**

We start by substituting \( y = x + \rho \) into the ellipse equation:

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

Expanding the second term:

\[ \frac{x^2}{a^2} + \frac{x^2 + 2x\rho + \rho^2}{b^2} = 1. \]

Combine the terms:

\[ \frac{x^2}{a^2} + \frac{x^2}{b^2} + \frac{2x\rho}{b^2} + \frac{\rho^2}{b^2} = 1. \]

**Step 2: Combine like terms.**

Multiply through by \( a^2b^2 \) to eliminate the denominators:

\[ b^2 x^2 + a^2 x^2 + 2a^2 x \rho + a^2 \rho^2 = a^2b^2. \]

Combine the \( x^2 \) terms:

\[ (b^2 + a^2)x^2 + 2a^2 x \rho + (a^2 \rho^2 - a^2b^2) = 0. \]

**Step 3: Use the quadratic formula to find the roots.**

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

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

where \( A = b^2 + a^2 \), \( B = 2a^2 \rho \), and \( C = a^2 \rho^2 - a^2b^2 \).

Using the quadratic formula \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):

\[ x = \frac{-2a^2 \rho \pm \sqrt{(2a^2 \rho)^2 - 4(b^2 + a^2)(a^2 \rho^2 - a^2b^2)}}{2(b^2 + a^2)}. \]

**Step 4: Find the midpoint of P and Q.**

Let the roots be \( x_1 \) and \( x_2 \). The midpoint \( h \) of \( P \) and \( Q \) is given by:

\[ h = \frac{x_1 + x_2}{2} = \frac{-B}{2A} = \frac{-2a^2 \rho}{2(b^2 + a^2)} = \frac{-a^2 \rho}{b^2 + a^2}. \]

**Step 5: Calculate the corresponding y-coordinate.**

Using \( y = x + \rho \):

\[ k = h + \rho = \frac{-a^2 \rho}{b^2 + a^2} + \rho = \rho \left(1 - \frac{a^2}{b^2 + a^2}\right) = \rho \left(\frac{b^2}{b^2 + a^2}\right). \]

**Step 6: Substitute \( h \) and \( k \) into the line equation.**

Now we substitute \( h \) and \( k \) into the line equation \( a^2y = -b^2x \):

Substituting \( x = h \) and \( y = k \):

\[ a^2\left(\frac{b^2 \rho}{b^2 + a^2}\right) = -b^2\left(\frac{-a^2 \rho}{b^2 + a^2}\right). \]

Both sides simplify to:

\[ \frac{a^2b^2 \rho}{b^2 + a^2} = \frac{a^2b^2 \rho}{b^2 + a^2}. \]

This shows that the midpoint \( (h, k) \) lies on the line \( a^2y = -b^2x \).

### Conclusion

Thus, we have proven that the midpoint of points P and Q lies on the line \( a^2y = -b^2x \).