What is ((x+7)^2}{25}+\frac{(y-3)^2)/4 =1 ?

The solution to ((x+7)^2}{25}+\frac{(y-3)^2)/4 =1 is Ellipse with (h,k)=(-7,3),a=5,b=2

Popular >

((x+7)^2}{25}+\frac{(y-3)^2)/4 =1

\frac{( x + 7 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1

(h, k) = (- 7, 3), a = 5, b = 2

Solution steps

\frac{( x + 7 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1

Rewrite

\frac{( x + 7 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1

in the form of the standard ellipse equation

\frac{( x - ( - 7 ) ) ^{2}}{5 ^{2}} + \frac{( y - 3 ) ^{2}}{2 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 7, 3), a = 5, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 7, 3), a = 5, b = 2

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

16 x^{2} + 96 x + 25 y^{2} + 100 y - 156 = 0

3 x^{2} - 6 x + y^{2} + 4 y - 2 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

4 x^{2} + 25 y^{2} - 40 x + 150 y + 225 = 0

What is ((x+7)^2}{25}+\frac{(y-3)^2)/4 =1 ?
The solution to ((x+7)^2}{25}+\frac{(y-3)^2)/4 =1 is Ellipse with (h,k)=(-7,3),a=5,b=2