Question

Find the equation of tangents to the curve y = cos(x + y), –2π ≤ x ≤ 2π that are parallel to the line x + 2y = 0.

Answer 1

Let the point of contact of one of the tangents be (x₁, y₁). Then (x₁, y₁) lies on y = cos(x + y).

$$\therefore y_1=\cos (x_1+y_1).....(i)$$

Since the tangents are parallel to the line x + 2y = 0. Therefore
Slope of tangent at (x₁, y₁) = slope of line x + 2y = 0

$$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=-\frac{1}{2}$$

The equation of curve is y = cos(x + y).
Differentiating with respect to x,

$$\frac{dy}{dx}=-\sin (x+y)(1+\frac{dy}{dx})$$

$$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=-\sin (x_1+y_1)\{1+{\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}\}$$

$$\Rightarrow -\frac{1}{2}=-\sin (x_1+y_1)(1-\frac{1}{2})$$

$$\Rightarrow \sin (x_1+y_1)=1.....(ii)$$

Squaring (i) and (ii) then adding,

$${\cos }^2(x_1+y_1)+{\sin }^2(x_1+y_1)={y_1}^2+1$$

$$\Rightarrow {y_1}^2+1=1$$

$$\Rightarrow y_1=0$$

Put 

$$y_1=0$$ in (i) and (ii),

$$\cos x_1=0\text{ and }\sin x_1=1$$

$$\Rightarrow x_1=\frac{\pi }{2},-\frac{3\pi }{2}$$

Hence, the points of contact are 

$$(\frac{\pi }{2},0)\text{ and }(-\frac{3\pi }{2},0)$$

The slope of the tangent is $$-\frac{1}{2}$$.

Therefore, equation of tangents at 

$$(\frac{\pi }{2},0)\text{ and }(-\frac{3\pi }{2},0)$$ are $$y-0=-\frac{1}{2}(x-\frac{\pi }{2})\text{ and }y-0=-\frac{1}{2}(x+\frac{3\pi }{2})$$

$$\text{ or }2x+4y-\pi =0\text{ and }2x+4y+3\pi =0$$