What is ((x-3)^2)/(25)+((y-5)^2)/(16)=1 ?

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

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((x-3)^2)/(25)+((y-5)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

13 x^{2} + 4 y^{2} - 52 x - 24 y + 36 = 0

36 - 9 x^{2} - 4 y^{2} \geq 0

(x 3)^{2} + (y + 2)^{2} = 25

(\frac{( y - 8 )}{40})^{2} + (\frac{( x + 10 )}{211})^{2} = 1

1 = (\frac{x}{84.5})^{2} + (\frac{y}{33.5})^{2}

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