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Problem 283
283
The equation of a curve is \(xy = 12\) and the equation of a line \(l\) is \(2x + y = k\), where \(k\) is a constant.
In the case where \(k = 10\), one of the points of intersection is \(P(2, 6)\). Find the angle, in degrees correct to 1 decimal place, between \(l\) and the tangent to the curve at \(P\).
Solution
The gradient of the line \(l\) is found by rearranging \(2x + y = k\) to \(y = -2x + k\). Thus, the gradient is \(-2\).
For the curve \(xy = 12\), differentiate implicitly: \(x \frac{dy}{dx} + y = 0\).
At \(P(2, 6)\), substitute \(x = 2\) and \(y = 6\) into the differentiated equation:
\(2 \frac{dy}{dx} + 6 = 0\)
\(\frac{dy}{dx} = -\frac{6}{2} = -3\)
The gradient of the tangent at \(P\) is \(-3\).
The angle \(\theta\) between the line and the tangent is given by: