What is ((x+3)^2)/(49)+(y^2)/(36)=1 ?

The solution to ((x+3)^2)/(49)+(y^2)/(36)=1 is Ellipse with (h,k)=(-3,0),a=7,b=6

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((x+3)^2)/(49)+(y^2)/(36)=1

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

(h, k) = (- 3, 0), a = 7, b = 6

Solution steps

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{7 ^{2}} + \frac{( y - 0 ) ^{2}}{6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 0), a = 7, b = 6

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, 0), a = 7, b = 6

25 x^{2} + 16 y^{2} - 350 x + 96 y - 231 = 0

4 x^{2} + 3 y^{2} - 12 = 0

x^{2} + y^{2} = 6 y

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{169} = 1

25 x^{2} + 4 y^{2} + 50 x - 16 y = 59

What is ((x+3)^2)/(49)+(y^2)/(36)=1 ?
The solution to ((x+3)^2)/(49)+(y^2)/(36)=1 is Ellipse with (h,k)=(-3,0),a=7,b=6