If the equation of the parabola, whose vertex is at (5,4) and directrix is 3x+y-29=0, is x…

Question

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

Accepted Answer

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

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

Solutions · 1