Let the abscissae of the two points P and Q be the roots of `2x^2-rx+p=0` and the ordinates of P and Q be the roots of `x^2-sx-q=0`. If the equation of the circle described on PQ as diameter is `2(x^2 + y^2) - 11x – 14y – 22=0`, then `2r+s–2q+p` is equal to —

Answer

To solve the problem, we need to derive the values of \( r \), \( s \), \( p \), and \( q \) from the given equations and the equation of the circle. Let's break it down step by step.

### Step 1: Identify the roots of the equations

The abscissae of points \( P \) and \( Q \) are the roots of the equation:
\[ 2x^2 - rx + p = 0 \]
Let the roots be \( x_1 \) and \( x_2 \). By Vieta's formulas:
- Sum of roots: \( x_1 + x_2 = \frac{r}{2} \)
- Product of roots: \( x_1 x_2 = \frac{p}{2} \)

The ordinates of points \( P \) and \( Q \) are the roots of the equation:
\[ y^2 - sy - q = 0 \]
Let the roots be \( y_1 \) and \( y_2 \). Again, by Vieta's formulas:
- Sum of roots: \( y_1 + y_2 = s \)
- Product of roots: \( y_1 y_2 = -q \)

### Step 2: Write the equation of the circle

The equation of the circle described on \( PQ \) as diameter is given by:
\[ 2(x^2 + y^2) - 11x - 14y - 22 = 0 \]
We can rearrange this to the standard form:
\[ x^2 + y^2 - \frac{11}{2}x - 7y - 11 = 0 \]

### Step 3: Compare coefficients

From the general form of the circle equation:
\[ x^2 + y^2 - (x_1 + x_2)x - (y_1 + y_2)y + (x_1 x_2 + y_1 y_2) = 0 \]
we can identify:
- \( x_1 + x_2 = \frac{r}{2} \) implies \( -\frac{r}{2} = -\frac{11}{2} \) → \( r = 11 \)
- \( y_1 + y_2 = s \) implies \( -s = -7 \) → \( s = 7 \)
- \( x_1 x_2 + y_1 y_2 = -11 \) implies \( \frac{p}{2} - q = -11 \) → \( p - 2q = -22 \)

### Step 4: Solve for \( 2r + s - 2q + p \)

Now we can substitute the values we found:
- \( r = 11 \)
- \( s = 7 \)
- We have the equation \( p - 2q = -22 \).

We need to calculate:
\[ 2r + s - 2q + p \]
Substituting the known values:
\[ 2(11) + 7 - 2q + p = 22 + 7 - 2q + p \]
This simplifies to:
\[ 29 - 2q + p \]
Using \( p = 2q - 22 \) from the earlier equation:
\[ 29 - 2q + (2q - 22) = 29 - 22 = 7 \]

### Final Answer

Thus, the value of \( 2r + s - 2q + p \) is:
\[ \boxed{7} \]