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Question

The orthocentre of any triangle formed by three tangents of parabola lies on the __ of the parabola.

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Solution

Let parabola be y2=4ax

Tangents at P(t1),Q(t2) and R(t3) are

t1y=x+at21 - - - - - - (1)

t2y=x+at22 - - - - - - (2)

t3y=x+at23 - - - - - - (3)

Points of intersection of these tangents are

A(at1t2,a(t1+t2))

B(at2t3,a(t2+t3))

C(at3t1,a(t3+t1))

Orthocenter is point of intersection of altitudes

So altitude from vertex A is

ya(t1+t2)=t3(xat1t2) - - - - - - (4)

Similarly altitude from B is

ya(t2+t3)=t1(xat2t3) - - - - - - (5)

So the orthocenter is the point of intersection of (4) and (5)

Hence by solving (4) and (5), we get

(x,y)(a,a(t1+t2+t3)+at1t2t3), Which is on the directrix of the parabola.


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