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Question

The length of a focal chord of the parabola y2=4ax at a distance b from the vertex is c, then

A
2a2=bc
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B
a3=b2c
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C
ac=b2
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D
b2c=4a3
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Solution

The correct option is D b2c=4a3
Let one end of the focal chord be (at2,2at), then the length of the that chord is
c=a(t+1t)2(1)
Now equation of the focal chord will be
y0=(2at0at2a)(xa)
As we know that vertex coordinates of the parabola is (0,0)
Hence perpendicular distance from origin to the chord will be
b=2att211+4t2(t21)2=|2at|t2+1(2)
from (1)×(2)2, t will get eliminated and we get
b2c=4a3

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