Заданная система уравнений х^2 + у^2 = 2, х+|y| = a графически представляет собой 3 фигуры:
- окружность х^2 + у^2 = 2,
- прямую у = -х + а,
- прямую у = х - а.
Эти прямые взаимно перпендикулярны и чтобы было 2 решения, они должны касаться окружности каждая в одной точке.
Радиусы в точку касания параллельны прямым, но так как они идут из начала координат, то их уравнения у = х и у = -х.
Возьмём у = х и у = -х + а и приравняем: 2х = а, х =а/2, но и у = х = а/2.
Подставим ув уравнение окружности: (а²/4) + (а²/4) = 2, 2а² = 8,
а² = 8/2 = 4. Отсюда а = +-2.
ответ: наибольшее значение параметра а равно 2.
Пошаговое объяснение:
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