What is ((x+2)^2}{49}+\frac{(y-3)^2)/4 =1 ?

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

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((x+2)^2}{49}+\frac{(y-3)^2)/4 =1

\frac{( x + 2 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{4} = 1

(h, k) = (- 2, 3), a = 7, b = 2

Solution steps

\frac{( x + 2 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 7, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), a = 7, b = 2

x^{2} + 4 y^{2} - 6 x + 12 y - 11 = 0

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

- y = x^{2} + y^{2}

\frac{( x - 7 ) ^{2}}{81} + \frac{( y ) ^{2}}{9} = 1

4 - x^{2} - 4 y^{2} = 1

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