Assertion A: Orthocentre of the triangle formed by any three tangents to the parabola lies on the directrix of the parabola.
Reason R: The orthocentre of the triangle formed by the tangents at $t_{1}, t_{2}, t_{3}$ to the parabola $y^{2} = 4 a x$ is $(- a, a (t_{1} + t_{2} + t_{3} + t_{1} t_{2} t_{3}))$

Both A and R are true and R is the correct explanation of A

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Both A and R are true and R is not the correct explanation of A

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A is true but R is false

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A is false but R is true

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Solution

The correct option is B Both A and R are true and R is the correct explanation of A
$y = m_{1} x + \frac{a}{m_{1}}$
$y = m_{2} x + \frac{a}{m_{2}} e q^{n}$ at tangent.
$y = m_{3} x + \frac{a}{m_{3}}$
$C = (\frac{a}{m_{1} m_{2}}, \frac{a (m_{1} + m_{2})}{m_{1} m_{2}})$
$B = (\frac{a}{m_{1} m_{3}}, \frac{a (m_{1} + m_{3})}{m_{1}, m_{3}})$
$(y - \frac{a (m_{1} + m_{2})}{m_{1} m_{2}}) = \frac{- 1}{m_{3}} (x - \frac{a}{m_{1}, m_{2}}) - - (1)$
$(y - \frac{a (m_{1} + m_{3})}{m_{1} m_{3}}) = \frac{- 1}{m_{2}} (x - \frac{a}{m_{1} m_{3}}) - - (2)$
Subtracting (1) - (2)
$x + a = 0$
$m_{1} = \frac{1}{t_{1}}, m_{2} = \frac{1}{t_{2}}, m_{3} = \frac{1}{t_{3}}$
$y = a (\frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{1}, m_{2}, m_{3}})$
$y = a (t_{1} + t_{2} + t_{3} + t_{1} t_{2} t_{3})$

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