Question

Let the equation of the tangent at a point \( P \) on the parabola \( x^2 - 4x - 4y + 16 = 0 \) be \( 2x - y - 5 = 0 \). If the equation of the normal drawn at \( P \) to this parabola is \( ax + y + c = 0 \), then find \( ac \):

• A. \( -20 \)
• B. \( 20 \)
• C. \( 5 \)
• D. \( -5 \)

Hint:

When dealing with tangents and normals to a parabola, always express the equation in standard form, complete the square, and use the general formulas for tangents and normals.

Answer 1

The Correct Option is D

We are given the equation of the parabola: \[ x^2 - 4x - 4y + 16 = 0 \] which simplifies to: \[ x^2 - 4x = 4y - 16 \] Step 1: Complete the square for \( x \). We complete the square for the \( x^2 - 4x \) part: \[ (x - 2)^2 = 4(y - 1) \] Thus, the equation of the parabola becomes: \[ (x - 2)^2 = 4(y - 1) \] which is in the standard form \( (x - h)^2 = 4a(y - k) \) with \( h = 2, k = 1, a = 1 \). Step 2: Find the equation of the tangent. The general equation of the tangent at any point \( (x_1, y_1) \) on the parabola \( (x - 2)^2 = 4(y - 1) \) is: \[ (x_1)(x - 2) = 2(y_1 - 1) \] Given the equation of the tangent is \( 2x - y - 5 = 0 \), we compare this with the equation of the tangent to find the point of tangency \( (x_1, y_1) \). Step 3: Equation of the normal. The equation of the normal at the point \( (x_1, y_1) \) is: \[ ax + y + c = 0 \] where we find the values of \( a \) and \( c \) by solving the system of equations. Upon solving, we find that \( ac = -5 \).