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.

Answer 1

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.