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प्रश्न
The equation of the tangent to the curve given by x = a sin3t, y = bcos3t at a point where t = `pi/2` is
पर्याय
y = 1
y = 0
x = 0
x = 1
MCQ
उत्तर
y = 0
Explanation:
x = a sin3t
`(dx)/(dt)` = 3a sin2t. cos t
y = b cos3t
`(dy)/(dx) = 3b cos^2t. (- sin t)`
`(dy)/(dx) = (dy/dt)/(dx/(dt)) = (-3b cos^2t. sint)/(3a sin^2t. cos t) = - b/a cot t`
Put `x = pi/2`
`m = (dy)/(dx) = 0, x = a sin^3 (pi/2) = a`
and `y = b cos^3 (pi/2)` = 0
∴ Equation of tangent y – 0 = 0 (x – 0) ⇒ y = 0
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