Skip to content
dronacharya.ai
← JEE Main 2022 · 27 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 11 of 90

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 11

If the equation of the parabola, whose vertex is at (5,4)(5,4) and the directrix is 3x+y−29=03x+y-29=0, is x2+ay2+bxy+cx+dy+k=0x^2+ay^2+bxy+cx+dy+k=0, then a+b+c+d+ka+b+c+d+k is equal to

Choose one option
Answer and worked solution

Official answer: D

Focus. The foot of the perpendicular from V(5,4)V(5,4) to the directrix is (8,5)(8,5), so the focus is (2,3)(2,3).

Equation. 10[(x−2)2+(y−3)2]=(3x+y−29)210\left[(x-2)^2+(y-3)^2\right]=(3x+y-29)^2 gives x2+9y2−6xy+134x−2y−711=0x^2+9y^2-6xy+134x-2y-711=0.

Sum. 9−6+134−2−711=−5769-6+134-2-711=-576.

(D)

Topics:ParabolaDistance of a point from a line
← Previous questionNext question →