AP EAMCET · Maths · Straight Lines
The locus of the centroid of the triangle with vertices at \((a \cos \theta, a \sin \theta),(b \sin \theta,-b \cos \theta)\) and \((1,0)\) is (here, \(\theta\) is a parameter)
- A \((3 x+1)^2+9 y^2=a^2+b^2\)
- B \((3 x-1)^2+9 y^2=a^2-b^2\)
- C \((3 x-1)^2+9 y^2=a^2+b^2\)
- D \((3 x+1)^2+9 y^2=a^2-b^2\)
Answer & Solution
Correct Answer
(C) \((3 x-1)^2+9 y^2=a^2+b^2\)
Step-by-step Solution
Detailed explanation
Given, vertices of a triangle are \(A(a \cos \theta, a \sin \theta), B(b \sin \theta, b \cos \theta)\) and \(C(1,0)\) \(\therefore\) Let the locus of centroid be \((x, y)\).…
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