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Question

A tangent is drawn at the point P(α,β) on the parabola y2=4ax. This tangent meets a second parabola y2=4a(x+k) at A and B, what is the midpoint of AB?


A

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B

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C

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D

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Solution

The correct option is D


Lets first see how the given situation looks like

Here, for the tangent to intersect parabola y2=4a(x+k) note that k should be positive. This is because the second parabola should be outside the first one, i.e, the vertex of the second parabola should lie on the left of the first parabola.

The tangents equation can be given by,

ty=x+at2 - - - - - - (1)

where (α,β)=(at2,2at)

Solving (1) and parabola y2=4a(x+k) we gets,

y2=4a(x+k)

=4a(ytat2+k)

i.e., y2+y(4at)+4a2t24ab=0

sum of roots = 4at

if A(x1,y1)B(x2,y2)

y1+y2=4at - - - - - - (2)

Midpoint of AB(x1+x22,y1+y22)

yMP=4at2=2at(using(2))

=β

Also,

[xt+at2t]2=4a(x+k)

x2+a2t4+2xat2=4at2x+kt2

x2+x(2at24at2)+a2t4kt2=0

x2+x(2at2)+a2t4kt2=0

sum of the roots=x1+x2=2at2

x1+x22=at2=α

midpoint of AB=(α,β)


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