If (xr,yr):r=1,2,3,4 be the points of intersection of the parabola y2=4ax and the circle x2+y2+2gx+2fy+c=0, then
A
y1+y2+y3+y4=0
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B
√x1+√x2+√x3+√x4=0
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C
y1−y2+y3−y4=0
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D
y1+y2−y3−y4=0
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Solution
The correct option is B√x1+√x2+√x3+√x4=0 Intersection points are (xr,yr):r=1,2,3,4 y2=4ax and the circle is x2+y2+2gx+2fy+c=0
From parabola equation, we have x=y24a
Now, y416a2+y2+2gy24a+2fy+c=0
Above equation has four roots.
As coefficient of y3 is 0,
so, y1+y2+y3+y4=0