The maximum distance from origin of a point on the curve x=a t-b ( a t b ) and y= t-b ( a…
Mathematics · JEE Main · NTA Exams — Co-ordinate Geometry
The maximum distance from origin of a point on the curve \(x=a \sin t-b \sin \left(\frac{a t}{b}\right)\) and \(y=\cos t-b \cos \left(\frac{a t}{b}\right) .\) both a, b > 0 is,
- \(\text { a - b }\)
- \(\mathrm{a}+\mathrm{b}\)
- \(\sqrt{a^{2}+b^{2}}\)
- \(\sqrt{a^{2}-b^{2}}\)
Answer
(B) a + b
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