The equation of a parabola is $25 [{(x - 2)}^{2} + {(y + 5)}^{2}] = {[3 x + 4 y - 1]}^{2}$ for this parabola

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Tangent at the vertex is $6 x + 8 y + 13 = 0$

Focus = $(2, - 5)$

Directrix has the equation $3 x + 4 y - 1 = 0$

Axis has the equation $4 x - 3 y - 23 = 0$

answer is A, B, C, D.

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Detailed Solution

${(x - 2)}^{2} + {(y + 5)}^{2} = {[\frac{3 x + 4 y - 1}{\sqrt{25}}]}^{2}$
$\begin{array}{l} S P^{2} = P M^{2} \\ s = (2, - 5) \\ d i r e c t r i x e q u a t i o n i s 3 x + 4 y - 1 = 0 \\ e q u a t i o n o f a x i s i s 4 x - 3 y + k = 0 \\ T h i s i s p a s s (2, - 5) \\ 8 + 15 + k = 0 \Rightarrow k = - 23 \\ ∴ E q u a t i o n i s 4 x - 3 y - 23 = 0 \\ T a n g e n t e q u a t i o n a t v e r t e x i s 3 x + 4 y + k = 0 \\ S Z = \frac{|6 - 20 - 1|}{\sqrt{9 + 16}} = \frac{|- 1 - 14|}{5} = \frac{15}{5} = 3 \\ \begin{matrix} n o w S A = \frac{3}{2} \\ \frac{|6 - 20 + k|}{5} = \frac{3}{2} \end{matrix} |\begin{matrix} \frac{|k - 14|}{5} = \frac{3}{2} \\ |k - 14| = \frac{15}{2} \end{matrix} \end{array}$
$\begin{array}{l} k - 14 = \pm \frac{15}{2} \\ k = 14 + \frac{15}{2} k = 14 - \frac{15}{2} \\ k = \frac{28 + 15}{2} k = \frac{28 - 15}{2} \\ k = \frac{43}{2} k = \frac{13}{2} \end{array}$
Tangent at vertex is $3 x + 4 y + \frac{13}{2}$
$6 x + 8 y + 13 = 0$