What is ((x+6)^2)/7+((y-3)^2)/(16)=1 ?

The solution to ((x+6)^2)/7+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(-6,3),a=sqrt(7),b=4

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((x+6)^2)/7+((y-3)^2)/(16)=1

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

(h, k) = (- 6, 3), a = 7, b = 4

Solution steps

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 6 ) ) ^{2}}{( 7 ) ^{2}} + \frac{( y - 3 ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 6, 3), a = 7, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 6, 3), b = 4, a = 7

4 x^{2} + y^{2} + 64 x - y + 201 = 0

5 x^{2} + 25 y^{2} - 125 = 0

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

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

64 x^{2} + y^{2} = 64

What is ((x+6)^2)/7+((y-3)^2)/(16)=1 ?
The solution to ((x+6)^2)/7+((y-3)^2)/(16)=1 is Ellipse with (h,k)=(-6,3),a=sqrt(7),b=4