Parabola Angle Calculation: \(\angle PAQ\)

This problem involves finding the angle formed at the vertex of a parabola, specifically the angle \(\angle PAQ\), where P and Q are points on the parabola and A is its vertex.

Understanding the Parabola and its Points

The equation of the parabola is given as \(y^2 = 4kx\).

• The standard form of a parabola with its vertex at the origin and opening horizontally is \(y^2 = 4ax\). By comparing this to the given equation, we see that \(4a = 4k\), which implies \(a = k\).
• The vertex A of this parabola is located at the origin, so its coordinates are A = (0, 0).
• The two points given are P(4k, 4k) and Q(4k, -4k).

Verifying Points Lie on the Parabola

Let's check if points P and Q satisfy the parabola's equation \(y^2 = 4kx\).

• For Point P(4k, 4k): Substitute \(x = 4k\) and \(y = 4k\) into the equation.
  • Left side: \(y^2 = (4k)^2 = 16k^2\).
  • Right side: \(4kx = 4k(4k) = 16k^2\).
  • Since \(16k^2 = 16k^2\), point P lies on the parabola.
• For Point Q(4k, -4k): Substitute \(x = 4k\) and \(y = -4k\) into the equation.
  • Left side: \(y^2 = (-4k)^2 = 16k^2\).
  • Right side: \(4kx = 4k(4k) = 16k^2\).
  • Since \(16k^2 = 16k^2\), point Q lies on the parabola.

Calculating Angle \(\angle PAQ\) using Slopes

We can determine the angle \(\angle PAQ\) by finding the slopes of the line segments AP and AQ. If two lines are perpendicular, the product of their slopes is -1.

• The formula for the slope (\(m\)) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
• Slope of AP (\(m_{AP}\)): Using A=(0, 0) and P=(4k, 4k). \( m_{AP} = \frac{4k - 0}{4k - 0} = \frac{4k}{4k} = 1 \)
• Slope of AQ (\(m_{AQ}\)): Using A=(0, 0) and Q=(4k, -4k). \( m_{AQ} = \frac{-4k - 0}{4k - 0} = \frac{-4k}{4k} = -1 \)
• Now, let's find the product of the slopes: \( m_{AP} \times m_{AQ} = 1 \times (-1) = -1 \)
• Since the product of the slopes is -1, the line segments AP and AQ are perpendicular.
• Therefore, the angle \(\angle PAQ = 90^\circ\).

Calculating Angle \(\angle PAQ\) using Vectors

Alternatively, we can use vectors to find the angle between AP and AQ.

• Represent the line segments as vectors originating from the vertex A(0, 0).
  • Vector \(\vec{AP} = P - A = (4k, 4k) - (0, 0) = \langle 4k, 4k \rangle\).
  • Vector \(\vec{AQ} = Q - A = (4k, -4k) - (0, 0) = \langle 4k, -4k \rangle\).
• The angle \(\theta\) between two vectors \(\vec{u}\) and \(\vec{v}\) is given by the dot product formula: \( \cos \theta = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| ||\vec{v}||} \)
• Calculate the dot product of \(\vec{AP}\) and \(\vec{AQ}\): \( \vec{AP} \cdot \vec{AQ} = (4k)(4k) + (4k)(-4k) \) \( = 16k^2 - 16k^2 \) \( = 0 \)
• A dot product of 0 signifies that the vectors are orthogonal (perpendicular).
• Thus, the angle between vectors \(\vec{AP}\) and \(\vec{AQ}\) is \(90^\circ\).
• Therefore, \(\angle PAQ = 90^\circ\).

Conclusion

Both the slope method and the vector method show that the angle \(\angle PAQ\) is \(90^\circ\). This means the lines connecting the vertex A to points P and Q are perpendicular to each other.