What is the foci (2x-3)^2 ?

The foci (2x-3)^2 is (3/2 , 1/16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci (2x-3)^2

foci

(2 x - 3)^{2}

(\frac{3}{2}, \frac{1}{16})

Solution steps

Rewrite

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

in the standard form:

4 \cdot \frac{1}{16} (y - 0) = (x - \frac{3}{2})^{2}

(h, k) = (\frac{3}{2}, 0), p = \frac{1}{16}

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

p

from the center

(\frac{3}{2}, 0)

along the y-axis

(\frac{3}{2}, 0 + p)

= (\frac{3}{2}, 0 + \frac{1}{16})

Refine

(\frac{3}{2}, \frac{1}{16})

\frac{x ^{2}}{6}

y = x^{2} + 12 x + 26

(y - 3)^{2} = 4 (x + 2)

y = 5 x^{2} + 40 x + 79

x^{2} + 3 x - 2