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).

\[\therefore y_1 = \cos\left( x_1 + y_1 \right) . . . . . (i)\]

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

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

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

\[\frac{dy}{dx} = - \sin\left( x + y \right)\left( 1 + \frac{dy}{dx} \right)\]

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

\[ \Rightarrow - \frac{1}{2} = - \sin\left( x_1 + y_1 \right)\left( 1 - \frac{1}{2} \right)\]

\[ \Rightarrow \sin\left( x_1 + y_1 \right) = 1 . . . . . (ii)\]

Squaring (i) and (ii) then adding,

\[\cos^2 \left( x_1 + y_1 \right) + \sin^2 \left( x_1 + y_1 \right) = {y_1}^2 + 1 \]

\[ \Rightarrow {y_1}^2 + 1 = 1\]

\[ \Rightarrow y_1 = 0\]

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 

\[\left( \frac{\pi}{2}, 0 \right) \text { and } \left( - \frac{3\pi}{2}, 0 \right)\]

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

Therefore, equation of tangents at 

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

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