IIT JEE Advanced 2006 Paper 1

From the figure shown here, we have OV = √2 and OS = 2√2.
The vertex V is obtained as follows:
(V(OV)cos45°, (OV)sin 45°) = (√2

1

√2

,√2

1

√2

) = (1 , 1)
S = ((OS)cos45°, (OS)sin45°)
S = (2, 2)

The equation of directrix is
OV = VS
where
OV = √2
VS = √(2−1)2+(2−1)2 = √2
Now , a = √2 and therefore,
x + y = 0 (directrix)
Equation of parabola: we have P(x, y) and
PS = PM
√(x−2)2+(y−2)2 =

|x+y|

√2

(x+y)2 = 2((x–2)2+(y–2)2)
x2+y2 + 2xy = 2(x2+y2) + 16 – 8x – 8y
x2+y2 – 2xy = 8x + 8y – 16
(x–y)2 = 8(x + y – 2)