If `A={(x,y):y^2=4ax}:B={(x,y):y^2=-4ax};c={(x,y):x^2=4ay}andD={(x,y):x^2=-4ay}.`Find the number of points in `AnnBnnCnnD.`

Answer

To solve the problem of finding the number of points in the intersection \( A \cap B \cap C \cap D \), we will analyze each set and determine their intersections step by step.

### Step 1: Define the sets
1. **Set A**: \( A = \{(x,y) : y^2 = 4ax\} \)
- This represents a right-opening parabola.

2. **Set B**: \( B = \{(x,y) : y^2 = -4ax\} \)
- This represents a left-opening parabola.

3. **Set C**: \( C = \{(x,y) : x^2 = 4ay\} \)
- This represents an upward-opening parabola.

4. **Set D**: \( D = \{(x,y) : x^2 = -4ay\} \)
- This represents a downward-opening parabola.

### Step 2: Find intersections of the sets
To find \( A \cap B \cap C \cap D \), we need to find common points among these curves.

#### Intersection of A and B
- From Set A: \( y^2 = 4ax \)
- From Set B: \( y^2 = -4ax \)

Setting these equal gives:
\[ 4ax = -4ax \]
This implies:
\[ 8ax = 0 \]
Thus, \( x = 0 \) (since \( a \) is a constant and cannot be zero). Substituting \( x = 0 \) into either equation gives:
\[ y^2 = 0 \Rightarrow y = 0 \]
So, the only intersection point of A and B is \( (0, 0) \).

#### Intersection of C and D
- From Set C: \( x^2 = 4ay \)
- From Set D: \( x^2 = -4ay \)

Setting these equal gives:
\[ 4ay = -4ay \]
This implies:
\[ 8ay = 0 \]
Thus, \( y = 0 \) (again, since \( a \) is a constant). Substituting \( y = 0 \) into either equation gives:
\[ x^2 = 0 \Rightarrow x = 0 \]
So, the only intersection point of C and D is also \( (0, 0) \).

### Step 3: Combine the intersections
Since the only common point found in both intersections (A ∩ B) and (C ∩ D) is \( (0, 0) \), we conclude that:
\[ A \cap B \cap C \cap D = \{(0, 0)\} \]

### Conclusion
Thus, the number of points in \( A \cap B \cap C \cap D \) is **1**.