Application of Derivatives 1 Question 21
21. Find all the tangents to the curve $y=\cos (x+y)$, $-2 \pi \leq x \leq 2 \pi$, that are parallel to the line $x+2 y=0$.
$(1985,5 M)$
Integer Answer Type Question
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Solution:
- Given, $y=\cos (x+y)$
$\Rightarrow \quad \frac{d y}{d x}=-\sin (x+y) \cdot 1+\frac{d y}{d x}$
Since, tangent is parallel to $x+2 y=0$,
then slope $\frac{d y}{d x}=-\frac{1}{2}$
From Eq. (i), $-\frac{1}{2}=-\sin (x+y) \cdot 1-\frac{1}{2}$
$\Rightarrow \quad \sin (x+y)=1$, which shows $\cos (x+y)=0$.
$\therefore \quad y=0$
$$ \begin{aligned} \Rightarrow & x+y & =\frac{\pi}{2} \text { or } & -\frac{3 \pi}{2} \\ \therefore & x & =\frac{\pi}{2} \text { or } & -\frac{3 \pi}{2} \end{aligned} $$
Thus, required points are $\frac{\pi}{2}, 0$ and $-\frac{3 \pi}{2}, 0$
$\therefore$ Equation of tangents are
$$ \frac{y-0}{x-\pi / 2}=-\frac{1}{2} $$
and
$$ \frac{y-0}{x+3 \pi / 2}=-\frac{1}{2} \Rightarrow 2 y=-x+\frac{\pi}{2} $$
and
$$ 2 y=-x-\frac{3 \pi}{2} $$
$\Rightarrow \quad x+2 y=\frac{\pi}{2}$
and
$$ x+2 y=-\frac{3 \pi}{2} $$
are the required equations of tangents.
