Question

The point of intersection of a pair of normals of a parabola has abcissa greater than?

• A. 2a
• B. a
• C. 0
• D. -2a

Answer 1

Answer: (A)

2a

Answer 2

The abcissa of the point of intersection of two normals to the parabola $y^2 = 4ax$ at points with parameters $t_1$ and $t_2$ is given by $x = a(2 + t_1^2 + t_1t_2 + t_2^2)$. For distinct normals, i.e., $t_1 \neq t_2$, the term $t_1^2 + t_1t_2 + t_2^2$ is always strictly positive. Therefore, $x = a(2 + \text{positive value}) > a(2 + 0)$, which means $x > 2a$.