What is ((y+3)^2)/(25)+((x+4)^2)/(16)=1 ?

The solution to ((y+3)^2)/(25)+((x+4)^2)/(16)=1 is Ellipse with (h,k)=(-4,-3),a=4,b=5

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((y+3)^2)/(25)+((x+4)^2)/(16)=1

\frac{( y + 3 ) ^{2}}{25} + \frac{( x + 4 ) ^{2}}{16} = 1

(h, k) = (- 4, - 3), a = 4, b = 5

Solution steps

\frac{( y + 3 ) ^{2}}{25} + \frac{( x + 4 ) ^{2}}{16} = 1

Rewrite

\frac{( y + 3 ) ^{2}}{25} + \frac{( x + 4 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 4 ) ) ^{2}}{4 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 4, - 3), a = 4, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 4, - 3), b = 5, a = 4

3 x^{2} + 4 y^{2} - 16 = 0

x^{2} + 4 x + 2 y^{2} + 6 y + 1 = 0

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

20 x^{2} + 36 y^{2} = 405

\frac{x}{x ^{2} + y ^{2}} = \frac{3}{4}

What is ((y+3)^2)/(25)+((x+4)^2)/(16)=1 ?
The solution to ((y+3)^2)/(25)+((x+4)^2)/(16)=1 is Ellipse with (h,k)=(-4,-3),a=4,b=5