Answer
There are 2 points here satisfying the exercise, which are $(\pi/4,1)$ and $(-\pi/4,-1)$. The graph is shown below.
Work Step by Step
$$y=\tan x \hspace{2cm}(-\pi/2\lt x\lt\pi/2)$$
1) First, find the derivative of $y$: $$y'=\sec^2x=\frac{1}{\cos^2x}$$
2) Take the tangent $(t)$ with the slope $y'$ at point $A$ and also at point $A$, take the normal line $(n)$ with slope $s$.
Since the tangent and the normal line at a point is perpendicular to each other: $$s\times y'=-1$$ $$s=-\frac{1}{y'}=-\frac{1}{\frac{1}{\cos^2x}}=-\cos^2x$$
And because the normal $(n)$ is parallel to the line $y=-x/2$, their slopes are equal. In other words, $$s=-\frac{1}{2}$$ $$-\cos^2x=-\frac{1}{2}$$ $$\cos^2x=\frac{1}{2}$$ $$\cos x=\pm\frac{\sqrt2}{2}$$
3) For $\cos x=\sqrt2/2$, we have $x=\pm\pi/4$. Therefore,
- For $x=\pi/4$, $$y=\tan\frac{\pi}{4}=1$$
The normal to the curve at point $(\pi/4,1)$ is $$y-1=s(x-\frac{\pi}{4})$$ $$y-1=-\frac{1}{2}(x-\frac{\pi}{4})$$ $$y-1=-\frac{1}{2}x+\frac{\pi}{8}$$ $$y=-\frac{1}{2}x+\frac{\pi}{8}+1=-\frac{1}{2}x+\frac{\pi+8}{8}$$
- For $x=-\pi/4$, $$y=\tan\Big(-\frac{\pi}{4}\Big)=-1$$
The normal to the curve at point $(-\pi/4,-1)$ is $$y+1=s(x+\frac{\pi}{4})$$ $$y+1=-\frac{1}{2}(x+\frac{\pi}{4})$$ $$y+1=-\frac{1}{2}x-\frac{\pi}{8}$$ $$y=-\frac{1}{2}x-\frac{\pi}{8}-1=-\frac{1}{2}x-\frac{(\pi+8)}{8}$$
For $\cos x=-\sqrt2/2$, we cannot find any values of $x\in(-\pi/2,\pi/2)$.
4) Therefore, there are 2 points here satisfying the exercise, which are $(\pi/4,1)$ and $(-\pi/4,-1)$. The graph is shown below.
