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 $y^{2} = 4 a x$

Tangents at $P (t_{1}), Q (t_{2}) a n d R (t_{3})$ are

$t_{1} y = x + a t_{1}^{2}$ - - - - - - (1)

$t_{2} y = x + a t_{2}^{2}$ - - - - - - (2)

$t_{3} y = x + a t_{3}^{2}$ - - - - - - (3)

Points of intersection of these tangents are

$A (a t_{1} t_{2}, a (t_{1} + t_{2}))$

$B (a t_{2} t_{3}, a (t_{2} + t_{3}))$

$C (a t_{3} t_{1}, a (t_{3} + t_{1}))$

Orthocenter is point of intersection of altitudes

So altitude from vertex A is

$∴ y - a (t_{1} + t_{2}) = - t_{3} (x - a t_{1} t_{2})$ - - - - - - (4)

Similarly altitude from B is

$y - a (t_{2} + t_{3}) = - t_{1} (x - a t_{2} t_{3})$ - - - - - - (5)

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

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

$(x, y) \equiv (- a, a (t_{1} + t_{2} + t_{3}) + a t_{1} t_{2} t_{3})$ , Which is on the directrix of the parabola.

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