Find the foci of the curve $x^{2} - 6 x y + y^{2} - 10 x - 10 y - 19 = 0$ and also its directrices.

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Solution

The equation of the curve is:-
$x^{2} - 6 x y + y^{2} - 10 x - 10 y - 19 = 0$
The foci are given by
$\frac{{(x^{^{'}} - 3 y^{^{'}} - 5)}^{2} - {(- 3 x^{^{'}} + y^{^{'}} - 5)}^{2}}{0} = \frac{(X) (Y)}{3}$
Where, $X = x^{^{'}} - 3 y^{^{'}} - 5$
$Y = - 3 x^{^{'}} + y^{^{'}} - 5$
and $\frac{X Y}{5} = x^{^{'}} X + y^{^{'}} Y - 5 x^{^{'}} - 5 y^{^{'}} - 19$
The first pair gives either $X = Y$ or $X = - Y$
If $X = Y$ , then $x = y$ and $X = - Y = - 2 x - 5$
Putting in the second pair, we get
$\frac{{(2 x + 5)}^{2}}{- 3} = - x (4 x + 10) - 10 x - 19$ or
$x^{2} + 5 x + 4 = 0$ ; so $x = - 4$ or $- 1$
and $y = - 4$ or $- 1$
The foci are therefore $(- 1, - 1)$ and $(- 4, - 4)$
The directrices are the polars of the foci;-
$- 3 x - 3 y + 5 + 5 - 19 = 0$
and $3 x + 3 y + 40 - 19 = 0$
i.e., $x + y + 3 = 0$ and $x + y + 7 = 0$

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Find the foci of the curve x2−6xy+y2−10x−10y−19=0 and also its directrices.

Find the foci of the curve $x^{2} - 6 x y + y^{2} - 10 x - 10 y - 19 = 0$ and also its directrices.