The normal chord of the parabola `y^2 = 4ax` at a point whose ordinate is equal to abscissą subtends a right angle at the

Options

A. focus
B. vertex
C. ends of latus rectum
D. none

Answer

To solve the problem, we need to find the normal chord of the parabola \( y^2 = 4ax \) at a point where the ordinate (y-coordinate) is equal to the abscissa (x-coordinate). Let's denote this point as \( P(t, t) \), where \( t \) is the value of both the x and y coordinates.

### Step 1: Identify the point on the parabola
The parabola is given by the equation \( y^2 = 4ax \). If the ordinate is equal to the abscissa, we can substitute \( y = t \) and \( x = t \) into the equation of the parabola:

\[ t^2 = 4a(t) \]

### Step 2: Solve for \( t \)
Rearranging the equation gives us:

\[ t^2 - 4at = 0 \]

Factoring out \( t \):

\[ t(t - 4a) = 0 \]

This gives us two solutions:

1. \( t = 0 \)
2. \( t = 4a \)

Thus, the points on the parabola where the ordinate equals the abscissa are \( P(0, 0) \) and \( P(4a, 4a) \).

### Step 3: Find the normal at the point \( P(4a, 4a) \)
To find the equation of the normal at the point \( P(4a, 4a) \), we first need the slope of the tangent at this point. The slope of the tangent to the parabola \( y^2 = 4ax \) at a point \( (x_0, y_0) \) is given by:

\[ \text{slope of tangent} = \frac{dy}{dx} = \frac{2a}{y_0} \]

At \( P(4a, 4a) \):

\[ \text{slope of tangent} = \frac{2a}{4a} = \frac{1}{2} \]

The slope of the normal is the negative reciprocal of the slope of the tangent:

\[ \text{slope of normal} = -2 \]

### Step 4: Write the equation of the normal
Using the point-slope form of the equation of a line, the equation of the normal at \( P(4a, 4a) \) is:

\[ y - 4a = -2(x - 4a) \]

Simplifying this gives:

\[ y - 4a = -2x + 8a \]

\[ y = -2x + 12a \]

### Step 5: Find the points where the normal intersects the parabola
To find the points where this normal intersects the parabola again, we substitute \( y = -2x + 12a \) into the parabola equation \( y^2 = 4ax \):

\[ (-2x + 12a)^2 = 4ax \]

Expanding and simplifying:

\[ 4x^2 - 48ax + 144a^2 = 4ax \]

\[ 4x^2 - 52ax + 144a^2 = 0 \]

Dividing through by 4:

\[ x^2 - 13ax + 36a^2 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula:

\[ x = \frac{13a \pm \sqrt{(13a)^2 - 4 \cdot 1 \cdot 36a^2}}{2 \cdot 1} \]

Calculating the discriminant:

\[ (13a)^2 - 144a^2 = 169a^2 - 144a^2 = 25a^2 \]

So,

\[ x = \frac{13a \pm 5a}{2} \]

This gives us two solutions:

1. \( x = 9a \)
2. \( x = 4a \)

### Step 7: Find corresponding y-coordinates
For \( x = 9a \):

\[ y = -2(9a) + 12a = -18a + 12a = -6a \]

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

### Conclusion
The normal chord subtends a right angle at the point \( (4a, 4a) \) and intersects the parabola again at \( (9a, -6a) \).

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