What is ((x-3)^2)/(49)+((y-5)^2)/(16)=1 ?

The solution to ((x-3)^2)/(49)+((y-5)^2)/(16)=1 is Ellipse with (h,k)=(3,5),a=7,b=4

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((x-3)^2)/(49)+((y-5)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

(h, k) = (3, 5), a = 7, b = 4

Solution steps

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 3 ) ^{2}}{7 ^{2}} + \frac{( y - 5 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (3, 5), a = 7, b = 4

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (3, 5), a = 7, b = 4

\frac{x ^{2}}{64} + \frac{( y - 3 ) ^{2}}{9} = 1

4 x^{2} + z^{2} = 16

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{36} = 1

100 x^{2} + 64 y^{2} - 100 x - 1575 = 0

x^{2} + \frac{y ^{2}}{29} = 1

What is ((x-3)^2)/(49)+((y-5)^2)/(16)=1 ?
The solution to ((x-3)^2)/(49)+((y-5)^2)/(16)=1 is Ellipse with (h,k)=(3,5),a=7,b=4