[m]x=\sqrt{2}\cdot sin^3t[/m]
[m]y=\sqrt{2}\cdot cos^3t[/m]
[m]y(\frac{π}{4})=\sqrt{2}\cdot (cos\frac{π}{4})^3=\sqrt{2}\cdot (\frac{\sqrt{2}}{2})^3=\frac{1}{2}[/m]
[m]x(\frac{π}{4})=\sqrt{2}\cdot (sin \frac{π}{4})^3=\sqrt{2}\cdot (\frac{\sqrt{2}}{2})^3=\frac{1}{2}[/m]
[m]y`_{x}=\frac{y`_{t}}{x`_{t}}[/m]
[m]y`_(t)=(\sqrt{2}\cdot cos^3t)`=3\sqrt{2}\cdot cos^2t\cdot(cost)`=-3\sqrt{2}\cdot cos^2t\cdot sint[/m]
[m]x`_(t)=(\sqrt{2}\cdot sin^3t)`=3\sqrt{2}\cdot sin^2t\cdot(sin t)`=3\sqrt{2}\cdot sin^2t\cdot cost[/m]
[m]y`_(t_{o})=y`(\frac{π}{4})=-3\sqrt{2}\cdot (cos\frac{π}{4})^2\cdot (sin\frac{π}{4})=-\frac{3}{2}[/m]
[m]x`_(t_{o})=3\sqrt{2}\cdot (sin\frac{π}{4})^2\cdot (cos\frac{π}{4})=\frac{3}{2}[/m]
[m]y`_{x_{o}}=\frac{y`_{t_{o}}}{x`_{t_{o}}}=-1[/m]
Подставляем в уравнение:
y-y_(o)=y`_(x_(o))*(x-x_(o))
[m]y-\frac{1}{2})=-1(x-\frac{1}{2})[/m]
[m]y=-x+1[/m]
y-y_(o)=[red]- [/red]y`_(x_(o))*(x-x_(o)) - уравнение нормали к кривой y=y(x) в точке (x_(o);y_(o))
[m]y-\frac{1}{2})=1(x-\frac{1}{2})[/m]
О т в е т.
[b][m]y=-x+1[/m][/b]-уравнение касательной
[b][m]y=x[/m][/b]-уравнение нормали