Question

If the equation of the parabola, whose vertex is at $(5, 4)$ and the 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

• A. 575
• B. -575
• C. 576
• D. -576

Answer 1

Answer: (D)

-576

Answer 2

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

Let foot of perpendicular of $(5, 4)$ on directrix be $(x_1, y_1)$.

$\frac{x_1−5}{3}=\frac{y_1−4}{1}=\frac{−(−10)}{10}$

$∴ (x_1, y_1) = (8, 5)$

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

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

$(x−2)^2+(y−3)^2=\frac{(3x+y−29)^2}{10}$

$⇒10(x^2+y^2−4x−6y+13)=9x2^+y^2+841+6xy−58y−174x$

$⇒x^2+9y^2−6xy+134x−2y−711=0$

and given parabola

$x^2+ay^2+bxy+cx+dy+k=0$

$∴ a = 9, b = –6, c = 134, d = –2, k = –711$

$∴ a + b + c + d + k = –576$