Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/1328

1204 These are sufficient to determine uniquely the coefficients in the matrix, and from the coefficient of $$h^9$$ we derive the differential equation to the non-singular cubic

$\frac{u_5^2\mathrm{U}_9 - 3\mathrm{U}_7^2}{u_5^3}f + \frac{\mathrm{V}_8 + 2u_5^4}{u_5}\phi_1 - \frac{\mathrm{U}_7}{u_5}\phi + 3u_5\psi = 0,$

which becomes

$u_5^2\mathrm{U}_7\mathrm{U}_9 - \mathrm{V}_8^2 - 4\mathrm{U}_7^3 + u_5^4\mathrm{V}_8 = 0,$

and is identical with that previously found in (36).

And the matrix of the non-singular cubic is

$\mathrm{U}_7u_4^2 + \frac{\mathrm{U}_7^2 - \mathrm{V}_8\mathrm{L}_6 - \mathrm{U}_7\mathrm{L}_6^2}{u_5^2}u^4 + \mathrm{V}_8 + \mathrm{U}_7\mathrm{L}_6.$

26. The matrix of the equation to the tangents to the cubic from the temporary origin.

To find this equation, put $$\pi = m\xi_1$$, and form the discriminant of the resulting expression in $$\xi$$ as a binary quantic. Equating this to zero, replace $$m$$ by $$\pi/\xi$$. Since we need only the coefficient of the highest power of $$\pi$$, the simplest method of finding it is to put $$\xi = 0$$ in the matrix of the cubic, and form the discriminant of the resulting expression in $$\pi$$.

The matrix required is

$9b_3^3 - 12bc_2,$

or

$\left(\phi + \phi_1\mathrm{L}_6 + \phi_2\mathrm{L}_6^2\right)^2 - 4f\left(\psi + \psi_1\mathrm{L}_6 + psi_2\mathrm{L}_6^2 + psi_3\mathrm{L}_6^3 + psi_4\mathrm{L}_6^2\right)\quad.\quad.\quad(47),$

and the degree of this in $$\mathrm{L}_6$$ denotes the number of tangents which can be drawn.

The conditions that the cubic may be nodal or cuspidal are that this matrix, as a function of $$\mathrm{L}_6$$, may have a linear factor twice or three times repeated.

27. Case of nodal cubic.

In this case we still determine uniquely all the ratios of the coefficients, except $$\psi/f$$, by the condition that the coefficients of $$h$$ up to the seventh vanish identically, while the differential equation is found by equating the coefficient of $$h^8$$ to zero.

The further condition, to determine $$\psi/f$$, is (47) that the discriminant of

$\left(f\mathrm{L}_6^2 + \psi\mathrm{L}_6 - \mathrm{U}_7f\right)^2 - 4u_5^4f\left(f\mathrm{L}_6 + \psi\right)$

may vanish.

Writing $$k$$ or $$\psi/2f$$, and $$x$$ for $$\mathrm{L}_6 + k$$, this equation becomes

$x^4 - 2\left(k^2 + \mathrm{U}_7\right)x^2 - 4u_5^4x + \left(k^2 + \mathrm{U}_7\right)^2 - 4u_5^4k = 0\quad.\quad.\quad.\quad(48).$