If the tangents to the parabola y2=4ax at (x1,y1),(x2,y2) intersect at (x3,y3), then
A
x1,x3,x2 are in A.P.
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B
x1,x3,x2 are in G.P.
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C
y1,y3,y2 are in A.P.
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D
y1,y3,y2 are in G.P.
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Solution
The correct option is Cy1,y3,y2 are in A.P. Let (x1,y1)=(at21,2at1) (x2,y2)=(at22,2at2) (x3,y3)=(at1t2,a(t1+t2)) ∴x1⋅x2=at21⋅at22=(at1t2)2=x23
So, x1,x3,x2 are in G.P.
and y3=a(t1+t2)=12(2at1+2at2)=12(y1+y2)
So, y1,y3,y2 are in A.P.