What is ((x-5)^2)/4+((y+2)^2)/1 =1 ?

The solution to ((x-5)^2)/4+((y+2)^2)/1 =1 is Ellipse with (h,k)=(5,-2),a=2,b=1

Popular >

((x-5)^2)/4+((y+2)^2)/1 =1

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{1} = 1

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

Solution steps

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{1} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{1} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (5, - 2), a = 2, b = 1

\frac{8 x ^{2} - 12 x + 8 y ^{2} + 20 y + 17}{8} - 2 = 0

\frac{x ^{2}}{20} + \frac{y ^{2}}{25} = 1

16 x^{2} + 25 y^{2} - 32 x - 100 y - 284 = 0

9 x^{2} - 54 x + 4 y^{2} - 8 y - 59 = 0

\frac{( x + 5 ) ^{2}}{20} + \frac{( y - 6 ) ^{2}}{11} = 1

What is ((x-5)^2)/4+((y+2)^2)/1 =1 ?
The solution to ((x-5)^2)/4+((y+2)^2)/1 =1 is Ellipse with (h,k)=(5,-2),a=2,b=1