What is 4^2+4y^2-20x+40y=-5 ?

The solution to 4^2+4y^2-20x+40y=-5 is Parabola with (h,k)=(-79/20 ,-5),p= 5/4

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4^2+4y^2-20x+40y=-5

4^{2} + 4 y^{2} - 20 x + 40 y = - 5

(h, k) = (- \frac{79}{20}, - 5), p = \frac{5}{4}

Solution steps

4^{2} + 4 y^{2} - 20 x + 40 y = - 5

Rewrite

4^{2} + 4 y^{2} - 20 x + 40 y = - 5

in the standard form

4 \cdot \frac{5}{4} (x - (- \frac{79}{20})) = (y - (- 5))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{79}{20}, - 5), p = \frac{5}{4}

- x^{2} + 16 x + 32 y = 128

y^{2} - 4 x + 8 = 0

(y + 2)^{2} = (x + \frac{1}{4})

4 x^{2} - 8 x - y + 4 = 0

- 25 x^{2} + y 9^{2} = 225

What is 4^2+4y^2-20x+40y=-5 ?
The solution to 4^2+4y^2-20x+40y=-5 is Parabola with (h,k)=(-79/20 ,-5),p= 5/4