Question Detail

Building on the previous question (Q8), the slope of the tangent at (x₀, y₀) is m_t = 3x₀/y₀.The slope of the normal m_n is the negative reciprocal of the tangent slope: m_n = -1 / m_t = -y₀ / (3x₀).The equation of the normal line at (x₀, y₀) is y - y₀ = m_n (x - x₀). Substituting and simplifying gives y₀x + 3x₀y - 4x₀y₀ = 0.Using the point (1, 2) which is on the curve (from the previous tangent question), we substitute x₀ = 1, y₀ = 2:2x + 3(1)y - 4(1)(2) = 0 => 2x + 3y - 8 = 0. This matches Option C (rearranged).