What is ((x+3)^2)/4+((y-5)^2)/(25)=1 ?

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

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((x+3)^2)/4+((y-5)^2)/(25)=1

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

(h, k) = (- 3, 5), a = 2, b = 5

Solution steps

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

Rewrite

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 3, 5), a = 2, b = 5

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 3, 5), b = 5, a = 2

5 x^{2} + 9 y^{2} = 100

\frac{( x ^{2} )}{( 4 )} + \frac{( ( y - 1 ) ^{2} )}{( 9 )} < 1

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

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

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

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