What is ((x-5)/5)^2+((y+7)/(16))^2=1 ?

The solution to ((x-5)/5)^2+((y+7)/(16))^2=1 is Ellipse with (h,k)=(5,-7),a=5,b=16

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((x-5)/5)^2+((y+7)/(16))^2=1

(\frac{x - 5}{5})^{2} + (\frac{y + 7}{16})^{2} = 1

(h, k) = (5, - 7), a = 5, b = 16

Solution steps

(\frac{x - 5}{5})^{2} + (\frac{y + 7}{16})^{2} = 1

Rewrite

(\frac{x - 5}{5})^{2} + (\frac{y + 7}{16})^{2} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{5 ^{2}} + \frac{( y - ( - 7 ) ) ^{2}}{1 6 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 7), a = 5, b = 16

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 7), b = 16, a = 5

2 x^{2} + 4 y^{2} + 18 x + 24 y + 44 = 0

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

10 x^{2} + 40 x + 7 y^{2} + 70 y + 145 = 0

4 x^{2} + 4 y^{2} = 100 - 20 x + x^{2}

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

What is ((x-5)/5)^2+((y+7)/(16))^2=1 ?
The solution to ((x-5)/5)^2+((y+7)/(16))^2=1 is Ellipse with (h,k)=(5,-7),a=5,b=16