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Find the equations of the tangent and normal to the curve `x = a sin 3t, y = cos 2 t " at " t = (pi)/(4)` |
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Answer» We have `x = sin 3t rArr (dx)/(dt) = 3 cos 3t` and `y = cos 2 t rArr (dy)/(dt) = - 2 sin 2t` `:. (dy)/(dx) = ((.^(dy)//_(dt)))/((.^(dx)//_(dt))) = (-2 sin 2t)/(3 cos 3t)` `rArr [(dy)/(dx)]_(t = (pi)/(4)) = (-2 sin (2 xx (pi)/(4)))/(3 cos (3 xx (pi)/(4))) = (-2 sin (.^(pi)//_(2)))/(-3 cos (.^(pi)//_(4))) = (2 sqrt2)/(3)` When `t = (pi)/(4)`, we have `x = "sin" (3pi)/(4) = sin (pi - (pi)/(4)) = "sin" (pi)/(4) = (1)/(sqrt2) and y = "cos"(2pi)/(4) = "cos"(pi)/(2) = 0` `:.` point of contact of `P ((1)/(sqrt2), 0)` Equation of the tangent at the point P is given by `(y - 0)/(x - (1)/(sqrt2)) = (2 sqrt2)/(3)` `rArr (sqrt2y)/(sqrt2x -1) = (2 sqrt2)/(3)` `rArr 3 sqrt2y = 4x - 2 sqrt2` `rArr 2 sqrt2x - 3y - 2 = 0` Equation of the normal at the point P is given by `(y - 0)/(x - (1)/(sqrt2)) = (-3)/(2 sqrt2) rArr (sqrt2y)/(sqrt2x -1) = (-3)/(2 sqrt2)` `rArr 4y = - 3 sqrt2 x + 3 rArr 3 sqrt2 x + 4y - 3 = 0` |
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