What is ((x-3)^2)/9+((y-4)^2)/(16)=1 ?

The solution to ((x-3)^2)/9+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(3,4),a=3,b=4

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((x-3)^2)/9+((y-4)^2)/(16)=1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

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

Solution steps

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 3 ) ^{2}}{9} + \frac{( y - 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

\frac{( y + 3 ) ^{2}}{25} + \frac{( x + 4 ) ^{2}}{16} = 1

3 x^{2} + 4 y^{2} - 16 = 0

x^{2} + 4 x + 2 y^{2} + 6 y + 1 = 0

2 \cdot x^{2} + y^{2} = 6

What is ((x-3)^2)/9+((y-4)^2)/(16)=1 ?
The solution to ((x-3)^2)/9+((y-4)^2)/(16)=1 is Ellipse with (h,k)=(3,4),a=3,b=4