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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

9 x^{2} + 25 y^{2} - 36 x - 189 = 0

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2 x^{2} + 3 y^{2} + 8 x + 9 y - 10 = 0

2 x^{2} + 3 y^{2} - 8 x + 6 y + 5 = 0

3 x^{2} + y^{2} + 2 x - 6 y - 3 = 0

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