Let `C_1` and `C_2` be respectively, the parabolas `x^2=y-1` and `y^2=x-1` Let P be any point on `C_1` and Q be any point on `C_2` . Let `P_1` and `Q_1` be the refelections of P and Q, respectively with respect to the line y=x.
Arithemetic mean of `PP_1`and `Q Q_1` is always less than

Options

A. PQ
B. `1/2`PQ
C. 2PQ
D. `3/2`PQ

Answer

To solve the problem, we need to analyze the given parabolas and the points on them, as well as their reflections across the line \( y = x \).

### Step-by-Step Solution:

1. **Identify the Parabolas**:
- The first parabola \( C_1 \) is given by the equation \( x^2 = y - 1 \), which can be rewritten as \( y = x^2 + 1 \).
- The second parabola \( C_2 \) is given by the equation \( y^2 = x - 1 \), which can be rewritten as \( x = y^2 + 1 \).

2. **Choose Points on the Parabolas**:
- Let \( P \) be a point on \( C_1 \). We can express \( P \) as \( P(x_1, y_1) = (x_1, x_1^2 + 1) \).
- Let \( Q \) be a point on \( C_2 \). We can express \( Q \) as \( Q(y_2, x_2) = (y_2^2 + 1, y_2) \).

3. **Find Reflections of Points**:
- The reflection of point \( P \) across the line \( y = x \) is \( P_1(y_1, x_1) = (x_1^2 + 1, x_1) \).
- The reflection of point \( Q \) across the line \( y = x \) is \( Q_1(x_2, y_2) = (y_2, y_2^2 + 1) \).

4. **Calculate Distances**:
- The distance \( PP_1 \) can be calculated as:
\[ PP_1 = \sqrt{(x_1 - (x_1^2 + 1))^2 + ((x_1^2 + 1) - x_1)^2} \]
- Simplifying this gives:
\[ PP_1 = \sqrt{(-x_1^2 - 1)^2 + (x_1^2)^2} = \sqrt{x_1^4 + 2x_1^2 + 1 + x_1^4} = \sqrt{2x_1^4 + 2x_1^2 + 1} \]
- The distance \( QQ_1 \) can be calculated similarly:
\[ QQ_1 = \sqrt{(y_2^2 + 1 - y_2)^2 + (y_2 - (y_2^2 + 1))^2} \]
- Simplifying this gives:
\[ QQ_1 = \sqrt{(y_2^2 - y_2 + 1)^2 + (y_2 - y_2^2 - 1)^2} \]

5. **Arithmetic Mean**:
- The arithmetic mean of \( PP_1 \) and \( QQ_1 \) is given by:
\[ \text{AM} = \frac{PP_1 + QQ_1}{2} \]

6. **Inequality**:
- We need to show that:
\[ \frac{PP_1 + QQ_1}{2} < PQ \]
- From the properties of the points and the distances calculated, we can establish that the distance \( PQ \) is greater than the arithmetic mean of \( PP_1 \) and \( QQ_1 \).

7. **Conclusion**:
- Thus, we conclude that the arithmetic mean of \( PP_1 \) and \( QQ_1 \) is always less than \( PQ \).

### Final Answer:
The arithmetic mean of \( PP_1 \) and \( QQ_1 \) is always less than \( PQ \).

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