If the equation $(x - 1)^{2} + (y - 2)^{2} = k (x + y + 3)^{2}$ represents a parabola, then the value of $2 k + 6$ is

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Solution

Given : $(x - 1)^{2} + (y - 2)^{2} = k (x + y + 3)^{2}$
Clearly, $k > 0$
$(x - 1)^{2} + (y - 2)^{2} = \frac{2 k (x + y + 3)^{2}}{2}$
$\Rightarrow (x - 1)^{2} + (y - 2)^{2} = 2 k {(\frac{x + y + 3}{\sqrt{2}})}^{2}$
$\Rightarrow \sqrt{(x - 1)^{2} + (y - 2)^{2}} = \sqrt{2 k} ∣ ∣ ∣ \frac{x + y + 3}{\sqrt{2}} ∣ ∣ ∣$
By definition of parabola,
$S P = P M$
$\Rightarrow \sqrt{2 k} = 1$
$\Rightarrow 2 k = 1$
$\Rightarrow 2 k + 6 = 7$

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