I only really have time to do one more question rn so I'll leave the other two for others to answer, might come back in a few days if no one has replied by then.

1. Distance of PK via distance formula = sqrt[(x-2)^2 + (y-5)^2]

The distance of P from x=-1, if you just think about it logically, will always be a horizontal distance since x=-1 is a vertical line. So the distance will be the absolute value of x coordinate of P - x coordinate of x=-1, i.e. |x-(-1)| = |x+1|

Now you have sqrt[(x-2)^2 + (y-5)^2] = 2|x+1| (since distance from K is twice the distance from x=-1)

Square both sides:

(x-2)^2 + (y-5)^2 = 4(x+1)^2

x^2 - 4x + 4 + y^2 - 10y + 25 = 4(x^2 + 2x + 1)

x^2 - 4x + y^2 - 10y + 29 = 4x^2 + 8x + 4

-3x^2 - 12x + y^2 - 10y + 25 = 0

3x^2 + 12 - y^2 + 10y - 25 = 0

You could probably leave that as the equation, but using 4 unit techniques you can rearrange that into the basic equation of a hyperbola. I can't really give you a diagram here but it is a hyperbola centered at (-2,5).

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