Trace the parabolas :
$(4 x + 3 y + 15)^{2} = 5 (3 x - 4 y) .$

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Solution

${(4 x + 3 y + 15)}^{2} = 5 (3 x - 4 y) {(\frac{4 x + 3 y + 15}{5})}^{2} = \frac{5}{5} (\frac{3 x - 4 y}{5}) {(\frac{4 x + 3 y + 15}{5})}^{2} = (\frac{3 x - 4 y}{5}) Y = \frac{4 x + 3 y + 15}{5}, X = \frac{3 x - 4 y}{5} Y^{2} = X$

comparing with the standard equation of parabola

$4 a = 1 a = \frac{1}{4}$

Vertex can be found by solving $X = 0$ and $Y = 0$

$4 x + 3 y + 15 = 0 3 x - 4 y = 0$

solving both we get $x = - \frac{12}{5}, y = - \frac{9}{5}$

so the vertex is $(- \frac{12}{5}, - \frac{9}{5})$

Focus is found by solving $Y = 0$ and $X + a = 0$

$12 x - 16 y + 5 = 0 4 x + 3 y + 15 = 0$

solving both we get $x = - \frac{51}{20}, y = - \frac{8}{5}$

so the focus is $(- \frac{51}{20}, - \frac{8}{5})$

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