What is (x-0)^2=4(-5/12)(y-0) ?

The solution to (x-0)^2=4(-5/12)(y-0) is Parabola with (h,k)=(0,0),p=-5/12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-0)^2=4(-5/12)(y-0)

(x - 0)^{2} = 4 (- \frac{5}{12}) (y - 0)

(h, k) = (0, 0), p = - \frac{5}{12}

Solution steps

(x - 0)^{2} = 4 (- \frac{5}{12}) (y - 0)

Rewrite

(x - 0)^{2} = 4 (- \frac{5}{12}) (y - 0)

in the standard form:

4 (- \frac{5}{12}) (y - 0) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, 0), p = - \frac{5}{12}

s - 25 = \frac{45}{2} t - \frac{3}{4} t^{2} - 33 - 75

2 x^{2} - 7 y \geq 21

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

y^{2} - 6 x + 5 y - 11 = 0

6 x^{2} + 8 x^{2} + 12 x - 32 y = 178

What is (x-0)^2=4(-5/12)(y-0) ?
The solution to (x-0)^2=4(-5/12)(y-0) is Parabola with (h,k)=(0,0),p=-5/12