AP EAMCET · Maths · Straight Lines
Equation of the locus of the centroid of the triangle whose vertices are \((a \cos k, a \sin k),(b \sin k,-b \cos k)\) and \((1,0)\), where \(k\) is a parameter, is
- A \((1-3 x)^2+9 y^2=a^2+b^2\)
- B \((3 x-1)^2+9 y^2=2 a^2+2 b^2\)
- C \((3 x+1)^2+(3y)^2=2 a^2+2 b^2\)
- D \((3 x+1)^2+(3y)^2=3 a^2+3 b^2\)
Answer & Solution
Correct Answer
(A) \((1-3 x)^2+9 y^2=a^2+b^2\)
Step-by-step Solution
Detailed explanation
Let A, B ad C the vertices of triangle \(\begin{aligned} & \mathrm{A}=(a \cos k, a \sin k) \\ & \mathrm{B}=(b \sin k-b \cos k) \\ & \mathrm{C}=(1,0) \end{aligned}\) Let \(\mathrm{G}(x, y)\) be the centroid, \(\therefore \quad x=\frac{a \cos k+b \sin k+1}{3}\)…
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