BITSAT 2005 Solved Question Paper

Given equation of line is
x+y=0 ........(i)
and equation of circle is
x2+y2+4y=0 ........(ii)
On solving Eqs. (i) and (ii),
x2+(−x)2+4(−x)=0
⇒2x2−4x=0
⇒2x(x−2)=0
⇒x=0,2and y=0,−2
Now taking option (c)
i.e., y2=2x
at poing (0,0)⇒0=0
and at point (2,−2)
⇒(−2)2=2(2)⇒4=4
∴ option (c) is the correct answer.