What is (1-x)^2+(2-y)^2-(7-x)^2+y^2=40 ?

The solution to (1-x)^2+(2-y)^2-(7-x)^2+y^2=40 is Parabola with (h,k)=(19/6 ,1),p=-3/2

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(1-x)^2+(2-y)^2-(7-x)^2+y^2=40

(1 - x)^{2} + (2 - y)^{2} - (7 - x)^{2} + y^{2} = 40

(h, k) = (\frac{19}{6}, 1), p = - \frac{3}{2}

Solution steps

(1 - x)^{2} + (2 - y)^{2} - (7 - x)^{2} + y^{2} = 40

Rewrite

(1 - x)^{2} + (2 - y)^{2} - (7 - x)^{2} + y^{2} = 40

in the standard form

4 (- \frac{3}{2}) (x - \frac{19}{6}) = (y - 1)^{2}

Therefore parabola properties are:

(h, k) = (\frac{19}{6}, 1), p = - \frac{3}{2}

(x + 7)^{2} = 6 (y - 4)

(y - 6)^{2} = - 12 (x + 4)

9 x^{2} + 18 x - 3 = 36 r (x)

x - 5 y < 12 + x^{2}

X^{2} - 2 Y - 1 = 0

What is (1-x)^2+(2-y)^2-(7-x)^2+y^2=40 ?
The solution to (1-x)^2+(2-y)^2-(7-x)^2+y^2=40 is Parabola with (h,k)=(19/6 ,1),p=-3/2