What is 4y^2-5x+2y+1=0 ?

The solution to 4y^2-5x+2y+1=0 is Parabola with (h,k)=(3/20 ,-1/4),p= 5/16

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4y^2-5x+2y+1=0

4 y^{2} - 5 x + 2 y + 1 = 0

(h, k) = (\frac{3}{20}, - \frac{1}{4}), p = \frac{5}{16}

Solution steps

4 y^{2} - 5 x + 2 y + 1 = 0

Rewrite

4 y^{2} - 5 x + 2 y + 1 = 0

in the standard form

4 \cdot \frac{5}{16} (x - \frac{3}{20}) = (y - (- \frac{1}{4}))^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{20}, - \frac{1}{4}), p = \frac{5}{16}

What is 4y^2-5x+2y+1=0 ?
The solution to 4y^2-5x+2y+1=0 is Parabola with (h,k)=(3/20 ,-1/4),p= 5/16