Parabola

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=
‌‌

1

2

|x1y2+x2y3+x3y4+x4y1−(y1x2+y2x3+y3x4+y4x1)|
‌=‌

1

2

‌rvert‌16×4+1×(−4)+1×(−16)+16×16−(16×1+4×1+(−4)×16+(−16)(16)∣
‌=300‌ sq. units ‌