What is the foci ((x+2)^2)/(25)-((y-1)^2)/(144)=1 ?

The foci ((x+2)^2)/(25)-((y-1)^2)/(144)=1 is (11,1),(-15,1)

Popular >

foci ((x+2)^2)/(25)-((y-1)^2)/(144)=1

foci

\frac{( x + 2 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{144} = 1

(11, 1), (- 15, 1)

Solution steps

(h + c, k), (h - c, k)

Calculate Hyperbola properties

\frac{( x + 2 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{144} = 1 :

Right-left Hyperbola with

(h, k) = (- 2, 1), a = 5, b = 12

(- 2 + c, 1), (- 2 - c, 1)

Compute

c :

c = 5^{2} + 1 2^{2} : 13

(- 2 + 13, 1), (- 2 - 13, 1)

Refine

(11, 1), (- 15, 1)

\frac{x ^{2}}{29} - \frac{y ^{2}}{20} = 1

\frac{( x + 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{16} = 1

\frac{( x + 4 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{25} = 1

\frac{( y + 4 ) ^{2}}{9} - \frac{( x - 3 ) ^{2}}{25} = 1

\frac{x ^{2}}{4} - \frac{( y + 1 ) ^{2}}{16} = 1

What is the foci ((x+2)^2)/(25)-((y-1)^2)/(144)=1 ?
The foci ((x+2)^2)/(25)-((y-1)^2)/(144)=1 is (11,1),(-15,1)