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Definition of Hyperbola

lf t is par...

Question

lf $t$ is parameter, $A = (a sec t, b tan t)$ and $B = (- a tan t, b sec t)$ , $O = (0, 0)$ then the locus of the centroid of $Δ O A B$ is

A

9 x y = a b

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B

x y = 9 a b

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C

x^{2} - 9 y^{2} = a^{2} - b^{2}

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D

x^{2} - y^{2} = \frac{1}{9} (a^{2} - b^{2})

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Solution

The correct option is A $9 x y = a b$
Let the centroid be G,
Then
$G = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$
Here
$(x_{1}, y_{1}) = (a s e c t, b t a n t)$
$(x_{2}, y_{2}) = (- a t a n t, b s e c t)$
$(x_{3}, y_{3}) = (0, 0)$
Hence
$G = (\frac{a s e c t - a t a n t}{3}, \frac{b t a n t + b s e c t}{3})$
Or
$G (x, y) = (\frac{a s e c t - a t a n t}{3}, \frac{b t a n t + b s e c t}{3})$
Therefore
$x = \frac{a s e c t - a t a n t}{3}$
Or
$3 x = a s e c t - a t a n t$ ..(i)
And
$y = \frac{b s e c t + b t a n t}{3}$
Or
$3 y = b s e c t + b t a n t$ ...(ii)
Hence
$3 x \times 3 y = (a s e c t - a t a n t) (b s e c t + b t a n t)$
$9 x y = a b (s e c t - t a n t) (s e c t + t a n t)$
$9 x y = a b (s e c^{2} t - t a n^{2} t)$
$9 x y = a b (1)$
Or
$9 x y = a b$

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