What is the vertices 10y^2=22x ?

The vertices 10y^2=22x is (h,k)=(0,0),p= 11/20

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vertices 10y^2=22x

vertices

10 y^{2} = 22 x

(h, k) = (0, 0), p = \frac{11}{20}

Solution steps

10 y^{2} = 22 x

Rewrite

10 y^{2} = 22 x

in the standard form:

4 \cdot \frac{11}{20} (x - 0) = (y - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = \frac{11}{20}

2 x^{2} + 5 x - 7

x^{2} + 12 y = 0

- 4.9 t^{2} + 9.8 t

- x^{2} + 4 x + 3

y = x^{2} - 4 x - 12