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

x^{2} + 4 y^{2} - 4 x - 12 = 0

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

\frac{( x + 1 ) ^{2}}{3} + \frac{( y - 1 ) ^{2}}{6} = 4

\frac{( x - 4 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

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

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