Step-by-Step Solution
Key Concept: For a conic section, the slope of the normal at point P(x₁, y₁) is the negative reciprocal of the tangent's slope. If the tangent slope is m, then the normal slope is -1/m. The third normal here uses the relationship between the point coordinate and the conic's parameter to establish that the normal slope equals y₁/2.
Given that the tangent with slope $m_3 = \frac{k}{2}$ passes through $(x_1, y_1)$, we have $m_3 = \frac{y_1}{x_1} \cdot 2$ from the condition. Therefore, $\frac{k}{2} = \frac{y_1}{x_1} \cdot 2$, which gives $m_3 = \frac{y_1}{2}$ as the slope of the tangent at the given point.
Correct Answer: 2