What is ((X+2)^2)/9+((Y-1)^2)/(16)=1 ?

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

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((X+2)^2)/9+((Y-1)^2)/(16)=1

\frac{( X + 2 ) ^{2}}{9} + \frac{( Y - 1 ) ^{2}}{16} = 1

(h, k) = (- 2, 1), a = 3, b = 4

Solution steps

\frac{( X + 2 ) ^{2}}{9} + \frac{( Y - 1 ) ^{2}}{16} = 1

Rewrite

\frac{( X + 2 ) ^{2}}{9} + \frac{( Y - 1 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( X - ( - 2 ) ) ^{2}}{3 ^{2}} + \frac{( Y - 1 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 1), a = 3, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 1), b = 4, a = 3

\frac{x}{x ^{2} + y ^{2}} \leq \frac{1}{2}

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

a^{2} + 2 b^{2} = 1

- 25 x^{2} - 4 y^{2} + 100 = 4

\frac{( x - 6 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{100} = 1

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