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Problem 2. Find the equation of the line normal to y = 4 sec ( 2 x ) y=4 \sec (2 x) y = 4 sec ( 2 x ) x = π 8 x=\frac{\pi}{8} x = 8 π y + 4 2 = 8 2 ( x − π 8 ) y+4 \sqrt{2}=8 \sqrt{2}\left(x-\frac{\pi}{8}\right) y + 4 2 = 8 2 ( x − 8 π ) y − 4 2 = − 8 2 ( x − π 8 ) y-4 \sqrt{2}=-8 \sqrt{2}\left(x-\frac{\pi}{8}\right) y − 4 2 = − 8 2 ( x − 8 π ) y + 4 2 = − 1 8 2 ( x + π 8 ) y+4 \sqrt{2}=-\frac{1}{8 \sqrt{2}}\left(x+\frac{\pi}{8}\right) y + 4 2 = − 8 2 1 ( x + 8 π ) y − 4 2 = − 1 8 2 ( x − π 8 ) y-4 \sqrt{2}=-\frac{1}{8 \sqrt{2}}\left(x-\frac{\pi}{8}\right) y − 4 2 = − 8 2 1 ( x − 8 π ) y + 4 2 = 8 2 ( x − π 8 ) y+4 \sqrt{2}=8 \sqrt{2}\left(x-\frac{\pi}{8}\right) y + 4 2 = 8 2 ( x − 8 π ) y − 4 2 = − 8 2 ( x − π 8 ) y-4 \sqrt{2}=-8 \sqrt{2}\left(x-\frac{\pi}{8}\right) y − 4 2 = − 8 2 ( x − 8 π ) y + 4 2 = − 1 8 2 ( x + π 8 ) y+4 \sqrt{2}=-\frac{1}{8 \sqrt{2}}\left(x+\frac{\pi}{8}\right) y + 4 2 = − 8 2 1 ( x + 8 π ) y − 4 2 = − 1 8 2 ( x − π 8 ) y-4 \sqrt{2}=-\frac{1}{8 \sqrt{2}}\left(x-\frac{\pi}{8}\right) y − 4 2 = − 8 2 1 ( x − 8 π ) Browse the problem library