The curve with equation
\(6e^{2x} + ke^y + e^{2y} = c\),
where \(k\) and \(c\) are constants, passes through the point \(P\) with coordinates \((\ln 3, \ln 2)\).
- Show that \(58 + 2k = c\).
- Given also that the gradient of the curve at \(P\) is \(-6\), find the values of \(k\) and \(c\).
Solution
(i) Substitute \(x = \ln 3\) and \(y = \ln 2\) into the equation:
\(6e^{2\ln 3} + ke^{\ln 2} + e^{2\ln 2} = c\).
Simplify using \(e^{2\ln 3} = 9\), \(e^{\ln 2} = 2\), and \(e^{2\ln 2} = 4\):
\(6 \times 9 + k \times 2 + 4 = c\).
\(54 + 2k + 4 = c\).
\(58 + 2k = c\).
(ii) Differentiate the left-hand side with respect to \(x\):
\(\frac{d}{dx}(6e^{2x} + ke^y + e^{2y}) = 0\).
\(12e^{2x} + ke^y \frac{dy}{dx} + 2e^{2y} \frac{dy}{dx} = 0\).
Substitute \(x = \ln 3\), \(y = \ln 2\), and \(\frac{dy}{dx} = -6\):
\(12 \times 9 + 2k(-6) + 2 \times 4(-6) = 0\).
\(108 - 12k - 48 = 0\).
\(60 = 12k\).
\(k = 5\).
Substitute \(k = 5\) into \(58 + 2k = c\):
\(58 + 2 \times 5 = c\).
\(c = 68\).
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