(a) Given that \(\begin{vmatrix} 5 & 2 & -3 \\ -1 & k & 6
(a) Given that \(\begin{vmatrix} 5 & 2 & -3 \\ -1 & k & 6 \\ 3 & 9 & (k + 2) \end{vmatrix} = -207\), find the values of the constant k.
(b) The equation of a curve is \(x(y^{2} + 1) - y(x^{2} + 1) + 4 = 0\). Find the:
(i) gradient of the curve at any point (x, y).
(ii) equation of the tangent to the curve at the point (-1, -3).
Explanation
(a) \(\begin{vmatrix} 5 & 2 & -3 \\ -1 & k & 6 \\ 3 & 9 & (k + 2) \end{vmatrix} = - 207\)
\(5[(k)(k + 2) - 54] - 2[(-1)(k + 2) - 18] - 3[-9 - 3k] = - 207\)
\(5k^{2} + 10k - 270 + 2k + 4 + 36 + 27 + 9k = - 207\)
\(5k^{2} + 21k - 203 + 207 = 0\)
\(5k^{2} + 21k + 4 = 0\)
\(5k^{2} + 20k + k + 4 = 0\)
\(5k(k + 4) + 1(k + 4) = 0\)
\((5k + 1)(k + 4) = 0\)
\(k = -\frac{1}{5}; k = -4\)
(b) Equation of curve:
\(x(y^{2} + 1) - y(x^{2} + 1) + 4 = 0\)
(i) Differentiating, we have
\(x(2y\frac{\mathrm d y}{\mathrm d x}) + y^{2} + 1 - y(2x) - (x^{2} + 1)\frac{\mathrm d y}{\mathrm d x} = 0\)
\(\frac{\mathrm d y}{\mathrm d x}(2xy) - \frac{\mathrm d y}{\mathrm d x}(x^{2} + 1) + y^{2} + 1 - 2xy = 0\)
\(\frac{\mathrm d y}{\mathrm d x}(2xy - x^{2} - 1) = 2xy - y^{2} - 1\)
The gradient of the curve at (x, y)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{2xy - y^{2} - 1}{2xy - x^{2} - 1}\)
(ii) Gradient of the tangent at (-1, -3) is
\(\frac{\mathrm d y}{\mathrm d x} = \frac{2(-1)(-3) - (-3)^{2} - 1}{2(-1)(-3) - (-1)^{2} - 1}\)
\(\frac{\mathrm d y}{\mathrm d x} = \frac{-4}{4} = -1\)
Equation : \(\frac{y - (-3)}{x - (-1)} = -1\)
\(y + 3 = - x - 1\)
\(y = - x -4\)