What is (x-3)+((y+1)/2)^2=5 ?

The solution to (x-3)+((y+1)/2)^2=5 is Parabola with (h,k)=(8,-1),p=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x-3)+((y+1)/2)^2=5

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

(h, k) = (8, - 1), p = - 1

Solution steps

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

Rewrite

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

in the standard form:

4 (- 1) (x - 8) = (y - (- 1))^{2}

Therefore parabola properties are:

(h, k) = (8, - 1), p = - 1

\frac{1}{2} x + \frac{1}{3} yy = 3 y

y^{2} - 2 x = 5

x^{2} + 4 x + 8 y + 44 = 0

4 x^{2} + 40 x + 106 + y = 0

x^{2} + y - 2 = 0

What is (x-3)+((y+1)/2)^2=5 ?
The solution to (x-3)+((y+1)/2)^2=5 is Parabola with (h,k)=(8,-1),p=-1