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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

\frac{x ^{2}}{15} + \frac{y ^{2}}{64} = 1

1 = \frac{x ^{2}}{4} + \frac{y ^{2}}{9}

28 x^{2} + 64 y^{2} - 84 x - 49 = 0

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

9 x^{2} + 25 y^{2} - 54 x - 100 y - 44 = 0

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