What is (2x^2)/(25)+((y-3)^2)/(49)=1 ?

The solution to (2x^2)/(25)+((y-3)^2)/(49)=1 is Ellipse with (h,k)=(0,3),a= 5/(sqrt(2)),b=7

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(2x^2)/(25)+((y-3)^2)/(49)=1

\frac{2 x ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

(h, k) = (0, 3), a = \frac{5}{2}, b = 7

Solution steps

\frac{2 x ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

Rewrite

\frac{2 x ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{2} ) ^{2}} + \frac{( y - 3 ) ^{2}}{7 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 3), a = \frac{5}{2}, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 3), b = 7, a = \frac{5}{2}

2 x^{2} + y^{2} = 1

12 x^{2} + 4 y^{2} - 24 x - 4 y + 1 = 0

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

4 x^{2} + 5 y^{2} = 20

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 3 ) ^{2}}{81} = 1

What is (2x^2)/(25)+((y-3)^2)/(49)=1 ?
The solution to (2x^2)/(25)+((y-3)^2)/(49)=1 is Ellipse with (h,k)=(0,3),a= 5/(sqrt(2)),b=7