What is 25^2+16y^2-200x-160y+400=0 ?

The solution to 25^2+16y^2-200x-160y+400=0 is Parabola with (h,k)=(25/8 ,5),p= 25/8

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25^2+16y^2-200x-160y+400=0

2 5^{2} + 16 y^{2} - 200 x - 160 y + 400 = 0

(h, k) = (\frac{25}{8}, 5), p = \frac{25}{8}

Solution steps

2 5^{2} + 16 y^{2} - 200 x - 160 y + 400 = 0

Rewrite

2 5^{2} + 16 y^{2} - 200 x - 160 y + 400 = 0

in the standard form

4 \cdot \frac{25}{8} (x - \frac{25}{8}) = (y - 5)^{2}

Therefore parabola properties are:

(h, k) = (\frac{25}{8}, 5), p = \frac{25}{8}

x + y = y^{2}

x^{2} + 6 x - 2 y + 15 = 0

y \leq - x^{2} - 5 x - \frac{13}{4}

x^{2} + 3 x + 5 y - 5 = 0

y^{2} = 4 c (x + c)

What is 25^2+16y^2-200x-160y+400=0 ?
The solution to 25^2+16y^2-200x-160y+400=0 is Parabola with (h,k)=(25/8 ,5),p= 25/8