TS EAMCET · Maths · Straight Lines
If \(\mathrm{t}\) is parameter, \(\mathrm{A}=(\mathrm{a} \sec \mathrm{t}, \mathrm{b} \tan \mathrm{t}), \mathrm{B}=(-\mathrm{a} \tan \mathrm{t}, \mathrm{b} \sec \mathrm{t})\) and \(\mathrm{O}=(0,0)\) then the locus of the centroid of \(\triangle \mathrm{OAB}\) is
- A \(9 x y=a b\)
- B \(x y=9 a b\)
- C \(x^2-9 y^2=a^2-b^2\)
- D \(x^2-y^2=\frac{1}{9}\left(a^2-b^2\right)\)
Answer & Solution
Correct Answer
(A) \(9 x y=a b\)
Step-by-step Solution
Detailed explanation
Centroid of \(\triangle A B C=(x, y)\) then \(\begin{aligned} x & =\frac{a \sec t-a \tan t}{3} \\ \Rightarrow \quad 3 x & =a(\sec t-\tan t)... (i) \\ & y=\frac{b \tan t+b \sec t}{3} \\ \Rightarrow \quad & 3 y=b(\tan t+\sec t)...(ii)\end{aligned}\)…
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