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Centre of Hyperbola

For parabola ...

Question

For parabola $y^{2} = 4 a x$ consider three points A, B, C lying on it. If the centroid $△ A B C$ is $(h_{1}, k_{1})$ & centroid triangle formed by the point of intersection of tangents at A, B, C has coordinates $(h_{2}, k_{2})$ then which of the following is always true.

A

2 k_{1} = k_{2}

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B

k_{1} = k_{2}

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C

k_{1}^{2} = \frac{4 a}{3} (h_{1} + 2 h_{2})

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D

k_{1}^{2} = \frac{4 a}{3} (2 h_{1} + h_{2})

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Solution

The correct option is B $k_{1} = k_{2}$
Given,
$y^{2} = 4 a x$ is a parabola with three points $A, B, C$ on its .
Now the parametric representation of these points will be
$A \equiv (a t_{1}^{2}, 2 a t_{1}) B \equiv ({a t}_{2}^{2}, 2 a t_{2}) C \equiv ({a t}_{3}^{2}, 2 a t_{3})$
Now, the centroid of the triangle formed using $A, B, C$ as the vertices will have coordinates
$(h_{1}, k_{1}) \equiv (\frac{a (t_{1}^{2} + t_{2}^{2} + t_{3}^{2})}{3}, \frac{2 a (t_{1} + t_{2} + t_{3})}{3})$
Also , the equation of a normal at $(a t^{2}, 2 a t)$ is
$y + t x = 2 a t + a t^{3}$
So equation of normal at $A, B, C$ will be
$y + t_{1} x = 2 a t_{1} + a t_{1}^{3} y + t_{2} x = 2 a t_{2} + a t_{2}^{3} y + t_{3} x = 2 a t_{3} + a t_{3}^{3}$
So the intersection the tangents accurs at three points having coordinates:
$D \equiv (a t_{1} t_{2}, a (t_{1} + t_{2})) E \equiv (a t_{1} t_{3}, a (t_{1} + t_{3})) F = (a t_{2} t_{3}, a (t_{2} + t_{3}))$
Again the centroid of the triangle formed using $D, E, F$ will be
$(h_{2}, k_{2}) \equiv (\frac{a (t_{1} t_{2} + t_{1} t_{3} + t_{2} t_{3})}{3}, \frac{2 a (t_{1} + t_{2} + t_{3})}{3}) ∴ k_{1} = k_{2}$

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