Given equation of parabola is y2=16x Comparing it with y2=4ax we get, a=4 Let coordinates of end points of focal chord of a parabola (i) is, (at2,2at) and (‌
a
t2
,‌
−2a
t
) ‌=‌ its length ‌=√(at2−‌
a
t2
)2+(2at+‌
2a
t
)2‌ (given) ‌ ‌=25 ‌⇒√(4(t2−‌
1
t2
))2+(8(t+‌
1
t
))2=25 (putting a=4 ) ‌⇒16(t2−‌
1
t2
)2+64(t+‌
1
t
)2=625 ‌⇒16(t4+‌
1
t4
−2)+64(t2+‌
1
t2
+2)=625 Let t+‌
1
t
=u so that (t+‌
1
t
)2=u2 ‌⇒‌‌t2+‌
1
t2
+2=u2 ‌⇒‌‌t2+‌
1
t2
=u2−2 ‌⇒‌‌(t2+‌
1
t2
)2=(u2−2)2 ‌⇒‌‌t4+‌
1
t4
+2=u4+4−2u2 Putting these values in Eq. (ii) and solving then we get, t=2,t=‌
1
2
,t=−2,t=‌
−1
2
So, Points are A(16,16),B(1,4),C(1,−4),D(16,−16) ∴ Area of quadrilateral ABCD= ‌‌