Questions: A parabola y^2=12x passes through the point (3,6). Find the i. Equation of the tangent; ii. Equation of the normal; to the parabola at the point, Q.

Solution Steps

To solve this problem, we need to find the equations of the tangent and normal lines to the given parabola at the point \( Q = (3, 6) \).

1. Equation of the Tangent:

   • First, find the derivative of the parabola equation \( y^2 = 12x \) to get the slope of the tangent line at any point.
   • Evaluate the derivative at the point \( Q = (3, 6) \) to find the slope of the tangent line.
   • Use the point-slope form of a line equation to write the equation of the tangent line.

2. Equation of the Normal:

   • The slope of the normal line is the negative reciprocal of the slope of the tangent line.
   • Use the point-slope form of a line equation to write the equation of the normal line.

The derivative of the parabola \( y^2 = 12x \) was calculated, resulting in \( \frac{dy}{dx} = 0 \) at the point \( Q(3, 6) \). This indicates that the slope of the tangent line at this point is \( m_{\text{tangent}} = 0 \).

Using the point-slope form of the line equation, the equation of the tangent line at the point \( Q(3, 6) \) is given by: \[ y - 6 = 0 \implies y = 6 \]

Since the slope of the tangent line is \( 0 \), the slope of the normal line, which is the negative reciprocal of the tangent slope, is undefined. This means the normal line is vertical.

The equation of the normal line, being vertical and passing through \( Q(3, 6) \), is given by: \[ x = 3 \]

Final Answer