Question

If the focus of a parabola is (2, -1) and directrix has the equation x+y = 3, then the vertex is

• A. (0, 3)
• B. (-1, ½)
• C. (-1, 2)
• D. (2, -1)

Answer 1

Answer: (C)

(-1, 2)

Answer 2

Let Z = (h, k) by the foot of the perpendicular from focii S = (-2, 1) to the directrix, then

$\frac{h + 2}{1} = \frac{k - 1}{1}\mspace{6mu} = \mspace{6mu}\frac{- ( - 2 + 1 - 3)}{1^{2} + 1^{2}}\mspace{6mu}\left( \frac{h - x_{1}}{\mathcal{l}} = \frac{k - y}{m} = \mspace{6mu}\frac{(\mathcal{l}n_{1} + my_{1} + n)}{\mathcal{l}^{2} + m^{2}} \right)$

⇒ (h, k) = (0, 3)

∴ The vertex is the mid point of SZ = (-1, 2)