@abhirup
H2 = AB . Hence,it is a parabola.!!
So, A is correct.
I'm sorry about (1.5,1.5)....I made a mistake..The option is actually (1.5,-1.5)
Also the curve IS symmetrical about x=-y
MULTIPLE OPTIONS CORRECT
The equation x2 + 2xy + x - 3 - y + y2 = 0 represents a conic with
(A) eccentricity 1 (B) vertex (1.5,-1.5)
(C) axis as x + y = 0 (D) length of latus rectum 1
@abhirup
H2 = AB . Hence,it is a parabola.!!
So, A is correct.
I'm sorry about (1.5,1.5)....I made a mistake..The option is actually (1.5,-1.5)
Also the curve IS symmetrical about x=-y
MOST CONVINCING METHOD FOR B IS :
PARTIAL DIFFERENTIATE; ONCE WITH Y AND NEXT WID X...
SOLVE SIMULT EQN.... U GET UR VERTEX.....
NOW LENGTH OF LATUS RECTUM NEEDS 2 B CN.....
x^2 + 2xy + x - 3 - y + y^2 = 0
==parabola .
(x+y)2 = (y-x+3)
=> Y2 = 4aX [Y=x+y, X=y-x+3]
so, 4a=1 => a=1/4
axis is X-axis i.e. Y=0 => x+y=0
vertex will be the intersection point of x+y=0 and y-x=3.
so, ans-> B and C.