Given the ellipse $3 x^{2} - 12 x + 2 y^{2} + 12 y + 6 = 0$ , and if there exists a real number $k = a \sqrt{b} + c$ where a, b, c are positive integers, b is not square of an integer, such that the hyperbola $xy - 2 y + 3 x = k$ is tangent to the above ellipse at two points, then which is/are correct

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a is prime

a is even

c=6

$[\frac{b}{a}] = 3 ([.] GIF)$

answer is A, B, C, D.

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Detailed Solution

$3 {(x - 2)}^{2} + 2 {(y - 3)}^{2} = 243 u^{2} + 2 v^{2} = 24$

$(x - 2) (y + 3) = k - 6 uv = k - 6$
They touch $m_{1} = m_{2} \Rightarrow - \frac{3 u}{2 v} = - \frac{v}{u} \Rightarrow 3 u^{2} = 2 v^{2}$
$u = \pm 2, v = \pm \sqrt{6} ∴ k - 6 = 2 \sqrt{6} \Rightarrow k = 2 \sqrt{6} + 6$
$\begin{array}{l} a = 2 \\ b = 6 \\ c = 6 \end{array}$