Application Of Derivatives Ques 17
17. Find all the tangents to the curve $y=\cos (x+y)$, $-2 \pi \leq x \leq 2 \pi$, that are parallel to the line $x+2 y=0$.
$(1985,5 \mathrm{M})$
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Answer:
($x+2y= \frac {\pi}{2} \text {and} x+2y= \frac {-3 \pi}{2}$)
Solution:
Formula:
Equation of tangent and normal :
- Given, $y=\cos (x+y)$
$\Rightarrow \quad (\frac{d y}{d x})=-\sin (x+y) \cdot (1+\frac{d y}{d x}) ……(i)$
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.
BETA
