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Question

If PQR is an equilateral triangle inscribed in the auxiliary circle of the ellipse x2a2+y2b2=1(a>b) and P′Q′R′ is the corresponding triangle inscribed within the ellipse then centroid of the triangle P′Q′R′ lies at

A
centre of ellipse
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B
focus of ellipse
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C
between focus and centre on major axis
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D
None of these
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Solution

The correct option is B centre of ellipse
We have the ellipse :
x2a2+y2b2=1, a>b,

The auxiliary circle has equation: x2+y2=a2
Parametric coordinate on auxiliary circle is:(acosα,asinα)
Parametric coordinate of the same point when it is projected on the ellipse becomes:(acosα,bsinα)
According to figure:
We take one point P at an angle α from x axis.
Now as we know that O is centre of Δ, Points Q and R will be at a difference of ±2π/3 from P.

In Co-ordinate form:
P:(acosα,asinα)
Q:(acos(α+(2π/3)), asin(α+(2π/3)))
R:(acos(α(2π/3)),asin(α(2π/3)))
Similarly,
P :(acos(α),bsin(α))
Q: (acos(α+(2π/3)),bsin(α+(2π/3)))
R :(acos(α(2π/3)),bsin(α(2π/3)))

Formula for centroid :(x1+x2+x33,y1+y2+y33)

Therefore putting values for P,Q,R, we get:
Centroid: (0,0)

745799_117627_ans_5575ee673f9842f68a63c90e813ae0e6.png

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