Sa Circle With Fixed Center (3h, 3k) And Of Variable Radius Cuts The Rectangular Hyperbola X2 – Y2 = 9a2 at The Points A, B, C, D. The Locus Of The Centroid Of The Triangle ABC Is Given By

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Question

Sa circle with fixed center (3h, 3k) and of variable radius cuts the rectangular hyperbola x2 – y2 = 9a2 at the points A, B, C, D. The locus of the centroid of the triangle ABC is given by

Solution

Correct option is

(x – 2h)2 – (y – 2k)2 = a2

Let A, B, C and D have coordinate (xiyi), i = 1, 2, 3, 4, then

                             ….. (1)

And                       ….. (2)

Let the centroid of ΔABC be  , then 

 .     

So, from (1) and (2),

      

But (x4y4) lies on x2 – y2 = 9a2

So,

 

 is (– 2h)2 – (y – 2k)2 = a2     

SIMILAR QUESTIONS

Q1

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Q2

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Q3

If the portion of the asymptotes between center and the tangent at the vertex of hyperbola  in the third quadrant is cut by the line  being parameter, then

Q4

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Q5

A, B, C and D are the points of intersection of a circle and a rectangular hyperbola which have different centers. If AB passes through the center of the hyperbola, then CD passes through

Q6

If PQ is a double ordinate of the hyperbola  such the OPQ is an equilateral triangle, O being the centre of the hyperbola. Then the eccentricity e of the hyperbola satisfies –

Q7

A rectangular hyperbola passes through the points A(1, 1), B(1, 5) and C(3, 1). The equation of normal to the hyperbola at A(1, 1) is –

Q8

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Q9

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Q10

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2x – y = 5 is –