The equation of a curve is \(y = 4x^2 - kx + \frac{1}{2}k^2\) and the equation of a line is \(y = x - a\), where \(k\) and \(a\) are constants.
Given instead that \(a = -\frac{7}{2}\), find the values of \(k\) for which the line is a tangent to the curve.
Solution
Substitute \(a = -\frac{7}{2}\) into the line equation, giving \(y = x + \frac{7}{2}\).
Set the curve equal to the line: \(4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2}\).
Rearrange to form a quadratic equation: \(4x^2 - (k+1)x + \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).
For the line to be tangent, the discriminant must be zero: \((k+1)^2 - 4 \times 4 \times \left(\frac{1}{2}k^2 - \frac{7}{2}\right) = 0\).
Simplify: \((k+1)^2 - 2k^2 + 14 = 0\).
Further simplify to: \(7k^2 - 2k - 57 = 0\).
Factor or use the quadratic formula to solve: \((k-3)(7k+19) = 0\).
Thus, \(k = 3\) or \(k = \frac{19}{7}\).
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