Through the focus of the parabola `y^2=2px(p gt0)` a line is drawn which intersects the curve at `A(x_1,y_1) & B(x_2,y_2)`. Then `(-y_1y_2)/(x_1x_2)` equals:

Answer

To solve the problem step by step, we will analyze the given parabola and the line drawn through its focus.

### Step 1: Identify the focus of the parabola
The equation of the parabola is given as \( y^2 = 2px \). This is a standard form of a parabola that opens to the right. The focus of this parabola is located at \( \left( \frac{p}{2}, 0 \right) \).

**Hint:** Remember that for the parabola \( y^2 = 4ax \), the focus is at \( (a, 0) \).

### Step 2: Parametric representation of points A and B
Let the points where the line intersects the parabola be \( A(x_1, y_1) \) and \( B(x_2, y_2) \). We can express these points in terms of parameters \( t_1 \) and \( t_2 \):
- For point A: \( A(t_1) = (at_1^2, 2at_1) \)
- For point B: \( B(t_2) = (at_2^2, 2at_2) \)

Here, \( a = \frac{p}{4} \).

**Hint:** Use the parametric form of the parabola to express points in terms of parameters.

### Step 3: Slope of the line through points A and B
Since the line intersects the parabola at points A and B, the slope of the line through these points can be expressed as:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} \]
This simplifies to:
\[ \frac{2a(t_2 - t_1)}{a(t_2^2 - t_1^2)} = \frac{2(t_2 - t_1)}{t_2^2 - t_1^2} \]
Using the difference of squares, we can rewrite this as:
\[ \frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_2 + t_1} \]

**Hint:** The slope can be simplified using the difference of squares formula.

### Step 4: Establishing the relationship between \( t_1 \) and \( t_2 \)
Since the slope of the line through points A and B remains constant, we can equate the slopes derived from the points:
\[ \frac{2(t_2 - t_1)}{t_2^2 - t_1^2} = \text{slope of the line} \]
This leads to a relationship between \( t_1 \) and \( t_2 \).

**Hint:** Use the property of slopes to derive relationships between parameters.

### Step 5: Calculate \( -\frac{y_1y_2}{x_1x_2} \)
Now we need to find \( -\frac{y_1y_2}{x_1x_2} \):
\[ y_1 = 2at_1, \quad y_2 = 2at_2, \quad x_1 = at_1^2, \quad x_2 = at_2^2 \]
Substituting these into the expression gives:
\[ -\frac{(2at_1)(2at_2)}{(at_1^2)(at_2^2)} = -\frac{4a^2t_1t_2}{a^2t_1^2t_2^2} = -\frac{4t_1t_2}{t_1^2t_2^2} = -\frac{4}{t_1t_2} \]
Since \( t_2 = -\frac{1}{t_1} \), we can substitute:
\[ -\frac{4}{t_1 \left(-\frac{1}{t_1}\right)} = -\frac{4}{-1} = 4 \]

**Hint:** Substitute the relationships found earlier to simplify the expression.

### Final Answer
Thus, we conclude that:
\[ -\frac{y_1y_2}{x_1x_2} = 4 \]