What is ((x-6)^2)/(16)+((y+3))/(25)=1 ?

The solution to ((x-6)^2)/(16)+((y+3))/(25)=1 is Parabola with (h,k)=(6,22),p=-4/25

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x-6)^2)/(16)+((y+3))/(25)=1

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 )}{25} = 1

(h, k) = (6, 22), p = - \frac{4}{25}

Solution steps

\frac{( x - 6 ) ^{2}}{16} + \frac{y + 3}{25} = 1

Rewrite

\frac{( x - 6 ) ^{2}}{16} + \frac{y + 3}{25} = 1

in the standard form:

4 (- \frac{4}{25}) (y - 22) = (x - 6)^{2}

Therefore parabola properties are:

(h, k) = (6, 22), p = - \frac{4}{25}

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

x = y^{2} + 4 y - 1

(y + 2)^{2} = 16 (x + 1)

7 x^{2} + y = 5

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

What is ((x-6)^2)/(16)+((y+3))/(25)=1 ?
The solution to ((x-6)^2)/(16)+((y+3))/(25)=1 is Parabola with (h,k)=(6,22),p=-4/25