What is 25(x+2)^2+4(y-3)^2=100 ?

The solution to 25(x+2)^2+4(y-3)^2=100 is Ellipse with (h,k)=(-2,3),a=2,b=5

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25(x+2)^2+4(y-3)^2=100

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

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

Solution steps

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

Rewrite

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

(x + 5)^{2} + 4 (y - 4)^{2} = 16

(x^{2} - 6 x) + 4 (y^{2} + 2 y) = - 9

\frac{( x + 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

\frac{( ( x + 0 ) ^{2} )}{100} + \frac{( ( y + 0 ) ^{2} )}{49} = 1

2 x^{2} + 3 y^{2} + 9 x + 10 y - 3 = 0

What is 25(x+2)^2+4(y-3)^2=100 ?
The solution to 25(x+2)^2+4(y-3)^2=100 is Ellipse with (h,k)=(-2,3),a=2,b=5