What is (((x+3))/(16))^2+((y-5)^2)/(25)=1 ?

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

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(((x+3))/(16))^2+((y-5)^2)/(25)=1

(\frac{( x + 3 )}{16})^{2} + \frac{( y - 5 ) ^{2}}{25} = 1

(h, k) = (- 3, 5), a = 16, b = 5

Solution steps

(\frac{x + 3}{16})^{2} + \frac{( y - 5 ) ^{2}}{25} = 1

Rewrite

(\frac{x + 3}{16})^{2} + \frac{( y - 5 ) ^{2}}{25} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 3 ) ) ^{2}}{1 6 ^{2}} + \frac{( y - 5 ) ^{2}}{5 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 3, 5), a = 16, b = 5

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 3, 5), a = 16, b = 5

- 4 x^{2} - 7 y^{2} > - 28

\frac{1}{4} (x - 2)^{2} + \frac{1}{16} (y + 7)^{2} = 1

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

1 x^{2} + 3 y^{2} + 9 x + 9 y - 12 = 0

\frac{y ^{2}}{4} + \frac{x ^{2}}{16} = 1

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