What is ((x+2)^2)/(16)+((y-3)^2)/(49)=1 ?

The solution to ((x+2)^2)/(16)+((y-3)^2)/(49)=1 is Ellipse with (h,k)=(-2,3),a=4,b=7

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((x+2)^2)/(16)+((y-3)^2)/(49)=1

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

(h, k) = (- 2, 3), a = 4, b = 7

Solution steps

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

Rewrite

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{4 ^{2}} + \frac{( y - 3 ) ^{2}}{7 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 3), a = 4, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 3), b = 7, a = 4

\frac{( x - 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{16} = 1

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

3 X^{2} + 2 Y^{2} = 8

9 x^{2} + 36 y^{2} = 324

5 x^{2} + 9 y^{2} - 20 x - 18 y - 16 = 0

What is ((x+2)^2)/(16)+((y-3)^2)/(49)=1 ?
The solution to ((x+2)^2)/(16)+((y-3)^2)/(49)=1 is Ellipse with (h,k)=(-2,3),a=4,b=7