What is the vertices 4y^2+9y^2+32x-18y+37=0 ?

The vertices 4y^2+9y^2+32x-18y+37=0 is (h,k)=(-25/26 , 9/13),p=-8/13

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vertices 4y^2+9y^2+32x-18y+37=0

vertices

4 y^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

(h, k) = (- \frac{25}{26}, \frac{9}{13}), p = - \frac{8}{13}

Solution steps

4 y^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

Rewrite

4 y^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

in the standard form

4 (- \frac{8}{13}) (x - (- \frac{25}{26})) = (y - \frac{9}{13})^{2}

Therefore parabola properties are:

(h, k) = (- \frac{25}{26}, \frac{9}{13}), p = - \frac{8}{13}

x^{2} + 3 x + 2

y = + (x - 3)^{2}

3 (x - 6)^{2} + 1

f (x) = (x + 3)^{2} - 2

f (x) = x^{2} + 4 x + 6

What is the vertices 4y^2+9y^2+32x-18y+37=0 ?
The vertices 4y^2+9y^2+32x-18y+37=0 is (h,k)=(-25/26 , 9/13),p=-8/13