Given parabola, y2+4x+2y−8=0 ...(i) Point of intersection of latus rectum and axis of parabola is foci. Now, from Eq. (i), we get ⇒(y+1)2−1+4x−8=0 ⇒(y+1)2=−4x+9 ⇒(y+1)2=−4(x−
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) Compare with general standard form of parabola (y−k)2=−4a(x−h) where, foci =(−a,0) Here, 4a=+4 ⇒a=+1 Then, (x−