What is 25y^2+40x+20y+28=0 ?

The solution to 25y^2+40x+20y+28=0 is Parabola with (h,k)=(-3/5 ,-2/5),p=-2/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

25y^2+40x+20y+28=0

25 y^{2} + 40 x + 20 y + 28 = 0

(h, k) = (- \frac{3}{5}, - \frac{2}{5}), p = - \frac{2}{5}

Solution steps

25 y^{2} + 40 x + 20 y + 28 = 0

Rewrite

25 y^{2} + 40 x + 20 y + 28 = 0

in the standard form

4 (- \frac{2}{5}) (x - (- \frac{3}{5})) = (y - (- \frac{2}{5}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{3}{5}, - \frac{2}{5}), p = - \frac{2}{5}

x^{2} = - \frac{1}{3} y

(y + 5)^{2} = - \frac{1}{3} (x - 3)

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

(y - (- 19))^{2} = 2 (x - 1)

24 y = 8 x^{2}

What is 25y^2+40x+20y+28=0 ?
The solution to 25y^2+40x+20y+28=0 is Parabola with (h,k)=(-3/5 ,-2/5),p=-2/5