The equation of a line is \(y = mx + c\), where \(m\) and \(c\) are constants, and the equation of a curve is \(xy = 16\).
(a) Given that the line is a tangent to the curve, express \(m\) in terms of \(c\).
(b) Given instead that \(m = -4\), find the set of values of \(c\) for which the line intersects the curve at two distinct points.
Solution
(a) To find when the line is tangent to the curve, substitute \(y = mx + c\) into \(xy = 16\):
\(x(mx + c) = 16\)
\(mx^2 + cx - 16 = 0\)
For the line to be tangent, the discriminant of this quadratic must be zero:
\(b^2 - 4ac = 0\)
Here, \(a = m\), \(b = c\), \(c = -16\).
\(c^2 - 4(m)(-16) = 0\)
\(c^2 + 64m = 0\)
\(m = -\frac{c^2}{64}\)
(b) Substitute \(m = -4\) into the equation:
\(x(-4x + c) = 16\)
\(-4x^2 + cx - 16 = 0\)
For two distinct points, the discriminant must be positive:
\(b^2 - 4ac > 0\)
\(c^2 - 4(-4)(-16) > 0\)
\(c^2 - 256 > 0\)
\(c^2 > 256\)
\(c > 16\) or \(c < -16\)
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