P is a point on the parabola whose ordinate equals its abscissa. A normal is drawn to the parabola at P to meet it again at Q. If S is the focus of the parabola, then the product of the slopes of `SP` and `SQ` is

Answer

To solve the problem, we need to find the product of the slopes of the segments \( SP \) and \( SQ \) where \( P \) is a point on the parabola \( y^2 = 4ax \) such that the ordinate equals the abscissa.

### Step-by-Step Solution:

1. **Identify the Parabola**:
The equation of the parabola is given as \( y^2 = 4ax \). The focus \( S \) of this parabola is at the point \( (a, 0) \).

2. **Find the Point \( P \)**:
Since the ordinate equals the abscissa, we can denote the coordinates of point \( P \) as \( (y, y) \). For the parabola, we have:
\[ y^2 = 4ax \implies y^2 = 4a(y) \implies y^2 - 4ay = 0 \]
Factoring gives:
\[ y(y - 4a) = 0 \]
Thus, \( y = 0 \) (the vertex) or \( y = 4a \). Since we want a point on the parabola other than the vertex, we take \( y = 4a \). Therefore, the coordinates of point \( P \) are \( (4a, 4a) \).

3. **Find the Normal at Point \( P \)**:
The slope of the tangent at point \( P \) can be found using the derivative of the parabola. The slope of the tangent line at point \( P \) is given by:
\[ \text{slope of tangent} = \frac{dy}{dx} = \frac{2}{4a} = \frac{1}{2} \]
Hence, the slope of the normal line at \( P \) is the negative reciprocal:
\[ \text{slope of normal} = -2 \]

4. **Equation of the Normal**:
The equation of the normal line at point \( P(4a, 4a) \) is:
\[ y - 4a = -2(x - 4a) \implies y = -2x + 12a \]

5. **Find Intersection Point \( Q \)**:
To find point \( Q \), we need to find where this normal intersects the parabola again. Substitute \( y = -2x + 12a \) into the parabola's equation \( y^2 = 4ax \):
\[ (-2x + 12a)^2 = 4ax \]
Expanding and rearranging gives:
\[ 4x^2 - 48ax + 144a^2 = 4ax \implies 4x^2 - 52ax + 144a^2 = 0 \]
Dividing through by 4:
\[ x^2 - 13ax + 36a^2 = 0 \]
Using the quadratic formula:
\[ x = \frac{13a \pm \sqrt{(13a)^2 - 4 \cdot 1 \cdot 36a^2}}{2 \cdot 1} = \frac{13a \pm \sqrt{169a^2 - 144a^2}}{2} = \frac{13a \pm 5a}{2} \]
This gives:
\[ x = 9a \quad \text{or} \quad x = 4a \]
Since \( x = 4a \) corresponds to point \( P \), we take \( x = 9a \). Substituting back to find \( y \):
\[ y = -2(9a) + 12a = -18a + 12a = -6a \]
Thus, the coordinates of point \( Q \) are \( (9a, -6a) \).

6. **Calculate the Slopes**:
- Slope of \( SP \):
\[ \text{slope of } SP = \frac{4a - 0}{4a - a} = \frac{4a}{3a} = \frac{4}{3} \]
- Slope of \( SQ \):
\[ \text{slope of } SQ = \frac{-6a - 0}{9a - a} = \frac{-6a}{8a} = -\frac{3}{4} \]

7. **Find the Product of the Slopes**:
Now, we calculate the product of the slopes:
\[ \text{Product} = \left(\frac{4}{3}\right) \left(-\frac{3}{4}\right) = -1 \]

### Final Answer:
The product of the slopes of \( SP \) and \( SQ \) is \( -1 \).