Prove that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus.

Answer

To prove that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus, we will follow these steps:

### Step 1: Define the Parabola and the Point
The equation of the parabola is given by:
\[ y^2 = 4ax \]
We need to find a point on the parabola where the ordinate (y-coordinate) is equal to the abscissa (x-coordinate). Let \( x = y \). Substituting \( y \) for \( x \) in the parabola's equation gives:
\[ y^2 = 4ay \]
This simplifies to:
\[ y^2 - 4ay = 0 \]
Factoring out \( y \):
\[ y(y - 4a) = 0 \]
Thus, \( y = 0 \) or \( y = 4a \). The corresponding points are:
- For \( y = 0 \): \( (0, 0) \)
- For \( y = 4a \): \( (4a, 4a) \)

### Step 2: Find the Normal at the Point \( (4a, 4a) \)
To find the normal at the point \( (4a, 4a) \), we first need the slope of the tangent line at this point. The derivative of \( y^2 = 4ax \) gives:
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \]
At the point \( (4a, 4a) \):
\[ \frac{dy}{dx} = \frac{2a}{4a} = \frac{1}{2} \]
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -2 \]

### Step 3: Equation of the Normal Line
Using the point-slope form of the line equation, the equation of the normal line at \( (4a, 4a) \) is:
\[ y - 4a = -2(x - 4a) \]
Simplifying this:
\[ y - 4a = -2x + 8a \implies y = -2x + 12a \]

### Step 4: Find the Focus of the Parabola
The focus of the parabola \( y^2 = 4ax \) is located at \( (a, 0) \).

### Step 5: Find the Points of Intersection with the Normal
To find where the normal intersects the parabola again, we substitute \( y = -2x + 12a \) into the parabola's equation:
\[ (-2x + 12a)^2 = 4ax \]
Expanding and simplifying:
\[ 4x^2 - 48ax + 144a^2 = 4ax \implies 4x^2 - 52ax + 144a^2 = 0 \]
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:
\[ x = 9a \quad \text{or} \quad x = 4a \]
The corresponding y-coordinates are:
- For \( x = 9a \): \( y = -2(9a) + 12a = -6a \)
- For \( x = 4a \): \( y = 4a \)

Thus, the points of intersection are \( (4a, 4a) \) and \( (9a, -6a) \).

### Step 6: Show that the Normal Chord Subtends a Right Angle at the Focus
The slope of the line segment from the focus \( (a, 0) \) to point \( (4a, 4a) \) is:
\[ m_1 = \frac{4a - 0}{4a - a} = \frac{4a}{3a} = \frac{4}{3} \]
The slope of the line segment from the focus \( (a, 0) \) to point \( (9a, -6a) \) is:
\[ m_2 = \frac{-6a - 0}{9a - a} = \frac{-6a}{8a} = -\frac{3}{4} \]
Now, we check if the product of the slopes \( m_1 \) and \( m_2 \) equals -1:
\[ m_1 \cdot m_2 = \frac{4}{3} \cdot -\frac{3}{4} = -1 \]
This confirms that the normal chord subtends a right angle at the focus.

### Conclusion
Thus, we have proved that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus.

---