What is x^28+(y^2)/(7^2)=1 ?

The solution to x^28+(y^2)/(7^2)=1 is Ellipse with (h,k)=(0,0),a= 1/(2sqrt(2)),b=7

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x^28+(y^2)/(7^2)=1

x^{2} 8 + \frac{y ^{2}}{7 ^{2}} = 1

(h, k) = (0, 0), a = \frac{1}{2 2}, b = 7

Solution steps

x^{2} \cdot 8 + \frac{y ^{2}}{7 ^{2}} = 1

Rewrite

x^{2} 8 + \frac{y ^{2}}{7 ^{2}} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{1}{2 2}, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 7, a = \frac{1}{2 2}

\frac{x ^{2}}{24} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

\frac{( ( x + 2 ) ^{2} )}{9} + \frac{( ( y - 4 ) ^{2} )}{16} = 1

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

- 16 y + 52 = - 2 x^{2} - 8 x - y^{2}

What is x^28+(y^2)/(7^2)=1 ?
The solution to x^28+(y^2)/(7^2)=1 is Ellipse with (h,k)=(0,0),a= 1/(2sqrt(2)),b=7