Find the equation of a parabola, whose focus (-6, -6) and the vertex is (-2, -2).

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$x^{2} + x y + y^{2} + 32 x + 32 y + 76 = 0$

$\Rightarrow x^{2} - 2 x y + y^{2} + 32 x + 32 y = 0$

$x^{2} - 2 x y + y^{2} + 32 x + 32 y + 76 = 0$

$2 x^{2} - 2 x y + y^{2} + 32 x + 32 y + 76 = 0$

answer is D.

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Detailed Solution

Let the vertex be V and the focus be S

The equation of axis is $x - y = 0$ .

Let the point Q is the point of intersection of the axis and the directrix.

Clearly, V is the mid-point of Q and S. Then Q is $(2, 2) .$

As we know that the directrix is perpendicular to the axis of the parabola. So, the equation of the directrix is

$x + y - k = 0$

which is passing through $(2, 2)$

Therefore, $k = 4 .$

Hence, the equation of the directrix is

$x + y - 4 = 0$

Thus the equation of the parabola is

$\begin{array}{l} \sqrt{(x + 6)^{2} + (y + 6)^{2}} = (\frac{x + y - 4}{\sqrt{1 + 1}}) \\ \Rightarrow 2 ((x + 6)^{2} + (y + 6)^{2}) = (x + y - 4)^{2} \\ \Rightarrow 2 (x^{2} + y^{2} + 12 x + 12 y + 36) = (x^{2} + y^{2} + 16 + 2 x y + 8 x - 8 y) \end{array}$