Question

The equation of the normal drawn to the parabola \( y^2 = 6x \) at the point \( (24,12) \) is:

• A. \( 3x - y = 60 \)
• B. \( 4x + y = 108 \)
• C. \( 2x + y = 60 \)
• D. \( x - 2y = 0 \)

Hint:

For a parabola of the form \( y^2 = 4ax \), use the derivative to find the tangent slope and then take the negative reciprocal to determine the normal slope.

Answer 1

The Correct Option is B

The given equation of the parabola is: \[ y^2 = 6x. \] Step 1: Finding the slope of the normal
Differentiating both sides with respect to \( x \): \[ 2y \frac{dy}{dx} = 6. \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{6}{2y} = \frac{3}{y}. \] At the point \( (24,12) \): \[ \frac{dy}{dx} = \frac{3}{12} = \frac{1}{4}. \] The slope of the tangent at \( (24,12) \) is \( \frac{1}{4} \), so the slope of the normal is the negative reciprocal: \[ m_{{normal}} = -4. \] Step 2: Finding the equation of the normal
Using the point-slope form of a line equation: \[ y - y_1 = m (x - x_1). \] Substituting \( (x_1, y_1) = (24,12) \) and \( m = -4 \): \[ y - 12 = -4(x - 24). \] Expanding: \[ y - 12 = -4x + 96. \] \[ y = -4x + 108. \] Rearranging: \[ 4x + y = 108. \] Thus, the equation of the normal is: \[ 4x + y = 108. \]