What is ((x-14)^2)/3+((y-11)^2)/4 =1 ?

The solution to ((x-14)^2)/3+((y-11)^2)/4 =1 is Ellipse with (h,k)=(14,11),a=sqrt(3),b=2

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((x-14)^2)/3+((y-11)^2)/4 =1

\frac{( x - 14 ) ^{2}}{3} + \frac{( y - 11 ) ^{2}}{4} = 1

(h, k) = (14, 11), a = 3, b = 2

Solution steps

\frac{( x - 14 ) ^{2}}{3} + \frac{( y - 11 ) ^{2}}{4} = 1

Rewrite

\frac{( x - 14 ) ^{2}}{3} + \frac{( y - 11 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - 14 ) ^{2}}{( 3 ) ^{2}} + \frac{( y - 11 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (14, 11), a = 3, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (14, 11), b = 2, a = 3

x^{2} + 7 y^{2} = 12

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 1 ) ^{2}}{25} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} < 1

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{\frac{9}{25}} = 1

What is ((x-14)^2)/3+((y-11)^2)/4 =1 ?
The solution to ((x-14)^2)/3+((y-11)^2)/4 =1 is Ellipse with (h,k)=(14,11),a=sqrt(3),b=2