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Question

Length of the focal chords of the parabola y2=4ax at a distance p from the vertex is

A
2a2p
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B
a2p2
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C
4a3p2
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D
p2a
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Solution

The correct option is C 4a3p2

y2=4ax
Slope of OP= Slope of OQ
t2=1t1
P(at2,2at) & Q(at2,2at)
Let length of focal chord be C.
(at2at2)2+(2at+2at)2=C
a2(t21t2)2+(2a)2(t+1t)2=C
a2(t+1t)2(t1t)2+(2a)2(t+1t)2=C
 a2(t+1t)2[(t1t)2+4]=C
a2(t+1t)2(t+1t22+4)=C
a2(t+1t)2(t+1t)2=C
a(t+1t)2=C(1)
Equation of PQ
y+2at=2at+2a/tat2at2(xat2)
y+2at=2t2+2tt41t2(xat2)
y(t41)+2at(t41)=2(t2+1)t(xat2)
y(t2+1)(t21)+2at(t2+1)(t21)=2t(t2+1)(xat2)
y(t21)+2at(t21)=2t(xat2)
y(t21)+2at(t21)=2tx2att2
2txy(t21)2att22at2t+2at=0
2txy(t21)2at=0
Equation of focal chord is 2txy(t21)2at=0
Q=|A1x1+B1y1+C1|A21+B21 (0,0)
P=∣ ∣ ∣2at(2t)2+(t21)2∣ ∣ ∣=2at4t2+t4+12t2
=2att4+1+2t2=2at(t2+1)2=2att2+1
t2+1t=2ap(2)
t+1t=2aP
From (1), we get,
a(2aP)2=C
C=4a2P2.a
C=4a3P2
Length of focal chord is 4a3P2.

1189328_1219375_ans_080839c8e71f4b65921c3233cd249837.png

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