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The two equations are:

  1. x = 2 - 2y
  2. x² + 4y² = 4

The points where the first meets the second are the values of x and y that simultaneously satisfy both equations:

Substitute using (1) for x in (2) and solve the formed quadratic:

x² + 4y² = 4

→ (2 - 2y)² + 4y² = 4

→ 4 - 8y + 4y² + 4y² -4 = 0

→ 8y² - 8y = 0

→ 8y(y - 1) = 0

→ y = 0 or y = 1

Substituting these values back into (1) gives:

y = 0 → x = 2 - 2× 0 = 2 - 0 = 2

y = 1 → x = 2 - 2× 1 = 2 - 2 = 0

→ The line meets the curve at the points (2, 0) and (0, 1)

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7y ago
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7y ago

The points (0, 1) and (2, 0).

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Q: Where are the points of contact when the line x equals 2 -2y crosses the curve x squared plus 4y squared equals 4?
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