Question

find the tangent to the curve y=cos(x+y),-2pi<x<2pi that are parallel to the line x+2y=0.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Curve is

$\rm :\longmapsto\:y = cos(x + y) - - - (1) \: \forall \: - 2\pi < x < 2\pi$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} cos(x + y)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: sin(x + y) \: \dfrac{d}{dx}(x + y)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: sin(x + y) \: \bigg(1 + \dfrac{dy}{dx}\bigg)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: sin(x + y) - sin(x + y) \dfrac{dy}{dx}$

$\rm :\longmapsto\:\dfrac{dy}{dx} + sin(x + y) \dfrac{dy}{dx} = - \: sin(x + y)$

$\rm :\longmapsto\:\dfrac{dy}{dx}\bigg(1 + sin(x + y)\bigg ) = - \: sin(x + y)$

$\bf\implies \:\dfrac{dy}{dx} = - \: \dfrac{sin(x + y)}{1 + sin(x + y)} - - - (2)$

Now it is given that tangent is parallel to the line x + 2y = 0

$\rm :\implies\:Slope \: of \: tangent = Slope \: of \: line \: x + 2y = 0$

$\rm :\implies\:\dfrac{dy}{dx} = - \: \dfrac{1}{2}$

$\rm :\longmapsto\: - \: \dfrac{sin(x + y)}{1 + sin(x + y)} = - \: \dfrac{1}{2}$

$\rm :\longmapsto\:\: \dfrac{sin(x + y)}{1 + sin(x + y)} = \: \dfrac{1}{2}$

$\rm :\longmapsto\:2sin(x + y) = 1 + sin(x + y)$

$\rm :\longmapsto\:sin(x + y) = 1 - - - (2)$

Now, Squaring equation (1) and (2) and add, we get

$\rm :\longmapsto\: {cos}^{2}(x + y) + {sin}^{2}(x + y) = {1}^{2} + {y}^{2}$

$\rm :\longmapsto\:1 = 1 + {y}^{2}$

$\rm :\longmapsto\: {y}^{2} = 0$

$\bf\implies \:y = 0 - - - (3)$

On substituting y = 0, in equation (2), we get

$\rm :\longmapsto\:sinx = 1$

$\bf\implies \:x = \dfrac{\pi}{2}, \: - \: \dfrac{3\pi}{2} \: as \: - 2\pi < x < 2\pi$

So, point of contact of tangent to the curve is

$\rm :\longmapsto\:\bigg(\dfrac{\pi}{2} , \: 0\bigg) \: and \: \bigg( - \: \dfrac{3\pi}{2} , \: 0\bigg)$

And

$\rm :\longmapsto\:Slope \: of \: tangent = - \: \dfrac{1}{2}$

Therefore,

Equation of tangent using slope point form is given by

$\rm :\longmapsto\:y - 0 = - \: \dfrac{1}{2}\bigg(x - \dfrac{\pi}{2} \bigg)$

and

$\rm :\longmapsto\:y - 0 = - \: \dfrac{1}{2}\bigg(x + \dfrac{3\pi}{2} \bigg)$

can be further simplified as

$\rm :\longmapsto\:2y= - x + \dfrac{\pi}{2}$

and

$\rm :\longmapsto\:2y= - x - \dfrac{3\pi}{2}$

can be further simplified as

$\red{\rm :\longmapsto\:2x + 4y = \pi} \\ \\ and \\ \\ \red{\rm :\longmapsto\:2x + 4y = - \: 3\pi}$

Additional Information :-

1. Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined.

4. Two lines having slope M and m are parallel, iff M = m.

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.