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

The solution to ((x+3)^2)/(16)+((y-1)^2)/(25)=1 is Ellipse with (h,k)=(-3,1),a=4,b=5

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

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

(h, k) = (- 3, 1), a = 4, b = 5

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{4 ^{2}} + \frac{( y - 1 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 1), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 1), b = 5, a = 4

16 x^{2} + 36 y^{2} - 96 x + 36 y + 9 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{16} = 1

(2 x - 2)^{2} + (3 y - 6)^{2} = 36

\frac{x ^{2}}{12} + \frac{y ^{2}}{16} = 1

\frac{3}{2} u^{2} + \frac{1}{2} v^{2} = 4

What is ((x+3)^2)/(16)+((y-1)^2)/(25)=1 ?
The solution to ((x+3)^2)/(16)+((y-1)^2)/(25)=1 is Ellipse with (h,k)=(-3,1),a=4,b=5