AP EAPCET 20-Aug-2021 Shift 2 Paper

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−

9

4

)
Compare with general standard form of parabola
(y−k)2=−4a(x−h)
where, foci =(−a,0)
Here, 4a=+4
⇒a=+1
Then, (x−

9

4

)=−1
⇒x=−1+

9

4

=

5

4

and (y+1)=0⇒y=−1
∴ Focus =(

5

4

,−1)= Point of intersection