What is the foci y=5x^2-12 ?

The foci y=5x^2-12 is (0,-239/20)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci y=5x^2-12

foci

y = 5 x^{2} - 12

(0, - \frac{239}{20})

Solution steps

Rewrite

y = 5 x^{2} - 12

in the standard form:

4 \cdot \frac{1}{20} (y - (- 12)) = (x - 0)^{2}

(h, k) = (0, - 12), p = \frac{1}{20}

Parabola is symmetric around the y-axis and so the focus lies a distance

p

from the center

(0, - 12)

along the y-axis

(0, - 12 + p)

= (0, - 12 + \frac{1}{20})

Refine

(0, - \frac{239}{20})

y = 2 x^{2} - 4 x + 9

y = 2 (x - 5)^{2} - 8

y = + 2 x^{2} + 2

\frac{1}{2} (x - 3)^{2} + 4

- x^{2} + 3 x - 5