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Question

The area of cyclic quadrilateral OABC is?

A
243
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B
482
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C
126
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D
185
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Solution

The correct option is A 243
Solving given parabolas, we have
8(xa)=4x
x=2a3
Points of intersection are (2a3,±8a3)
Now OABC is concyclic.
Fig.3.133
Hence, OAB must be right angle.
Slope of OA× Slope of AB=1
8a32a3×8a3a2a3=1
a=12
Coordinates of A and B are (8,42) and (8,42) respectively
Length of common chord =82
Area of quadrilateral =12OB×AC
=12×12×82
=482
Tangent to parabola y2=4x at point (8,42) is 42y=2(x+8) or x22y+8=0 which meets the x -axis at
D(8,0)
Tangent to parabola y2=8(x12) at point (8,42) is
42y=4(x+8)+96 or x+2y16=0, which meets the x -axis at E(16,0)
Hence, area of quadrilateral DAEC=12DE×AC
=12×24×82
=962

1632888_873451_ans_55a5df40914c4d7f8f8c0687cb61c132.png

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