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Question
Find the equation of all the tangents to the curve y = cos(x + y), –2π ≤ x ≤ 2π, that are parallel to the line x + 2y = 0.
Solution
Given that y = cos(x + y)
⇒ `"dy"/"dx" = - sin(x + y) [1 + "dy"/"dx"]` ....(i)
or `"dy"/"dx" = - (sin(x + y))/(1 + sin(x + y))`
Since tangent is parallel to x + 2y = 0, therefore slope of tangent = `- 1/2`
Therefore, `- (sin(x + y))/(1 + sin(x + y)) = - 1/2`
⇒ sin(x + y) = 1 .....(ii)
Since cos(x + y) = y and sin(x + y) = 1
⇒ cos2(x + y) + sin2(x + y) = y2 + 1
⇒ 1 = y2 + 1 or y = 0.
Therefore, cosx = 0.
Therefore, x = `(2"n" + 1) pi/2`, n = 0, ± 1, ± 2...
Thus, x = `+- pi/2, +- (3pi)/2`, but x = `pi/2`, x = `(-3pi)/2` satisfy equation (ii)
Hence, the points are `(pi/2, 0), ((-3pi)/2, 0)`.
Therefore, equation of tangent at `(pi/2, 0)` is y = `- 1/2(x - pi/2)`
or 2x + 4y – π = 0, and equation of tangent at `((-3pi)/2, 0)` is y = `- 1/2(x + (3pi)/2)`
or 2x + 4y + 3π = 0.
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