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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 $$\Delta OAB$$ is


A
9xy=ab
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B
xy=9ab
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C
x29y2=a2b2
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D
x2y2=19(a2b2)
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Solution

The correct option is A $$9\mathrm{x}\mathrm{y}= ab$$
Let the centroid be G,
Then
$$G=(\dfrac{x_{1}+x_{2}+x_{3}}{3},\dfrac{y_{1}+y_{2}+y_{3}}{3})$$
Here
$$(x_{1},y_{1})=(asect,btant)$$
$$(x_{2},y_{2})=(-atant,bsect)$$
$$(x_{3},y_{3})=(0,0)$$
Hence
$$G=(\dfrac{asect-atant}{3},\dfrac{btant+bsect}{3})$$
Or
$$G(x,y)=(\dfrac{asect-atant}{3},\dfrac{btant+bsect}{3})$$
Therefore
$$x=\dfrac{asect-atant}{3}$$
Or
$$3x=asect-atant$$ ..(i)
And
$$y=\dfrac{bsect+btant}{3}$$
Or
$$3y=bsect+btant$$ ...(ii)
Hence
$$3x\times3y=(asect-atant)(bsect+btant)$$
$$9xy=ab(sect-tant)(sect+tant)$$
$$9xy=ab(sec^{2}t-tan^{2}t)$$
$$9xy=ab(1)$$
Or
$$9xy=ab$$

Maths

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