QJEEM22Ma0c93: JEE Main 2022 Mathematics - Class 11 JEEM22Ma0c93 Solution: If the equation of the parabola, whose vertex is at (5, 4) and…

Given vertex is (5, 4) and directrix 3x + y $-$ 29 = 0

Let foot of perpendicular of (5, 4) on directrix is (x₁, y₁)

${{{x_1} - 5} \over 3} = {{{y_1} - 4} \over 1} = {{ - ( - 10)} \over {10}}$

$\therefore$ $({x_1},\,{y_1}) \equiv (8,\,5)$

So, focus of parabola will be $S = (2,3)$

Let P(x, y) be any point on parabola, then

${(x - 2)^2} + {(y - 3)^2} = {{{{(3x + y - 29)}^2}} \over {10}}$

$\Rightarrow 10({x^2} + {y^2} - 4x - 6y + 13) = 9{x^2} + {y^2} + 841 + 6xy - 58y - 174x$

$\Rightarrow {x^2} + 9{y^2} - 6xy + 134x - 2y - 711 = 0$

and given parabola

${x^2} + a{y^2} + bxy + cx + dy + k = 0$

$\therefore$ a = 9, b = $-$ 6, c = 134, d = $-$ 2, k = $-$ 711

$\therefore$ $a + b + c + d + k = - 576$