JEE Main 2022 — Parabola Question with Solution

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

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

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

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

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

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

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

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

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

and given parabola

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

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

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