TS EAMCET 5 May 2018 Shift 2 Solved Paper

The directrix of the parabola is,
x+y+4=0
Let P(x,y) be any point on the parabola.
Then,
SP=PM
(SP)2=(PM)2
(x+1)2+(y+1)2=(

|x+y+4|

√12+12

)2
2(x2+y2+2x+2y+2) =x2+y2+2xy+8y+8x+16
Solve further.
x2+y2−4x−4y−2xy −12=0
(x−y)2−4x−4y−12=0
(x−y)2=4(x+y+3)
Let
X=x−y
Y=x+y+3
Therefore,
X2=4Y
Compare with X2=−4AY
A=−1
Therefore, X=0,Y=0
Hence,
x=−

3

2

,y=

−3

2