If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $3x + y - 29 = 0$, is ${x^2} + a{y^2} + bxy + cx + dy + k = 0$, then $a + b + c + d + k$ is equal to :

<p>Given vertex is (5, 4) and directrix 3x + y $-$ 29 = 0</p> <p>Let foot of perpendicular of (5, 4) on directrix is (x₁, y₁)</p> <p>${{{x_1} - 5} \over 3} = {{{y_1} - 4} \over 1} = {{ - ( - 10)} \over {10}}$</p> <p>$\therefore$ $({x_1},\,{y_1}) \equiv (8,\,5)$</p> <p>So, focus of parabola will be $S = (2,3)$</p> <p>Let P(x, y) be any point on parabola, then</p> <p>${(x - 2)^2} + {(y - 3)^2} = {{{{(3x + y - 29)}^2}} \over {10}}$</p> <p>$$ \Rightarrow 10({x^2} + {y^2} - 4x - 6y + 13) = 9{x^2} + {y^2} + 841 + 6xy - 58y - 174x$$</p> <p>$\Rightarrow {x^2} + 9{y^2} - 6xy + 134x - 2y - 711 = 0$</p> <p>and given parabola</p> <p>${x^2} + a{y^2} + bxy + cx + dy + k = 0$</p> <p>$\therefore$ a = 9, b = $-$6, c = 134, d = $-$2, k = $-$711</p> <p>$\therefore$ $a + b + c + d + k = - 576$</p>