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Question

Two equal parabolas with axes in opposite directions touch at a point O. From a point P on one of them are drawn tangents PQ and PQ' to the other. Prove that QQ' will touch the first parabola in P' where PP' is parallel to the common tangent at O.

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Solution

Let the two parabolas be

y2=4ax.......(i)y2=4ax......(ii)

Let tangents are drawn from a point P(at2,2at) on (i) to (ii)

Then QQ is a chord of contact to (ii) w.r.t. to P

Equation of chord of contact is T=0

yy1=2a(x+x1)2aty=2a(x+at2)ty=xat2......(iii)

substituting x from (iii) in (i)

y2=4a(at2ty)y2+4aty+4a2t2=0(y+2at)2=0y=2at

We are getting equal roots from the equation. So, QQ touch the parabola

Substituting y=2at in (i)

(2at)2=4axx=at2

So the point of contact to (i) is P(at2,2at)

Equation of common tangent at O is

x=0

Slope of PP=2at2atat2at2=

So, PP is perpendicular to x axis

Hence it is parallel to the y axis or x=0

Hence proved.


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