What is ((x-2)^2)/(25)+((y-4)^2)/(16)=1 ?

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

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((x-2)^2)/(25)+((y-4)^2)/(16)=1

\frac{( x - 2 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

(h, k) = (2, 4), a = 5, b = 4

Solution steps

\frac{( x - 2 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 4), a = 5, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (2, 4), a = 5, b = 4

9 x^{2} + 25 y^{2} - 54 x - 100 y + 44 = 0

\frac{( x + 1 ) ^{2}}{8} + \frac{( y - \frac{3}{2} ) ^{2}}{4} = 1

\frac{x ^{2}}{1.5869} + \frac{y ^{2}}{0.9988} = 1

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

4 x^{2} + y^{2} \leq 4

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