(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(2yfrac{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)