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Problem 378
378
The equation of a curve is \(y^2 + 2x = 13\) and the equation of a line is \(2y + x = k\), where \(k\) is a constant. In the case where \(k = 8\), find the coordinates of the points of intersection of the line and the curve.
Solution
Given the equations:
1. \(y^2 + 2x = 13\)
2. \(2y + x = 8\)
From equation (2), express \(x\) in terms of \(y\):
\(x = 8 - 2y\)
Substitute \(x = 8 - 2y\) into equation (1):
\(y^2 + 2(8 - 2y) = 13\)
Simplify:
\(y^2 + 16 - 4y = 13\)
\(y^2 - 4y + 3 = 0\)
Factor the quadratic equation:
\((y - 3)(y - 1) = 0\)
Thus, \(y = 3\) or \(y = 1\).
For \(y = 3\):
\(x = 8 - 2(3) = 2\)
Point: \((2, 3)\)
For \(y = 1\):
\(x = 8 - 2(1) = 6\)
Point: \((6, 1)\)
Therefore, the points of intersection are \((2, 3)\) and \((6, 1)\).