What is (x-4)^2=-4(-5)(y+2) ?

The solution to (x-4)^2=-4(-5)(y+2) is Parabola with (h,k)=(4,-2),p=5

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

(x - 4)^{2} = - 4 (- 5) (y + 2)

(h, k) = (4, - 2), p = 5

Solution steps

(x - 4)^{2} = - 4 (- 5) (y + 2)

Rewrite

(x - 4)^{2} = - 4 (- 5) (y + 2)

in the standard form:

4 \cdot 5 (y - (- 2)) = (x - 4)^{2}

Therefore parabola properties are:

(h, k) = (4, - 2), p = 5

\frac{1}{t + 5} = \frac{6}{r ^{2} - 25}

y^{2} - 8 y + 8 x + 16 = 0

25 = \frac{x}{y ^{2}}

x^{2} - 76 x + 20 y + 50 = 0

x 0.14 x - 0.03 y = 0.48

What is (x-4)^2=-4(-5)(y+2) ?
The solution to (x-4)^2=-4(-5)(y+2) is Parabola with (h,k)=(4,-2),p=5