What is 4x^2+16y^2-64=1 ?

The solution to 4x^2+16y^2-64=1 is Ellipse with (h,k)=(0,0),a=(sqrt(65))/2 ,b=(sqrt(65))/4

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4x^2+16y^2-64=1

4 x^{2} + 16 y^{2} - 64 = 1

(h, k) = (0, 0), a = \frac{65}{2}, b = \frac{65}{4}

Solution steps

4 x^{2} + 16 y^{2} - 64 = 1

Rewrite

4 x^{2} + 16 y^{2} - 64 = 1

in the form of the standard ellipse equation

\frac{( x - 0 ) ^{2}}{( \frac{65}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{65}{4} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (0, 0), a = \frac{65}{2}, b = \frac{65}{4}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = \frac{65}{2}, b = \frac{65}{4}

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

\frac{( x - 6 ) ^{2}}{36} + \frac{y ^{2}}{100} = 1

\frac{2 y}{x ^{2} + y ^{2}} < 1

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

\frac{x ^{2}}{6 ^{2}} + \frac{y ^{2}}{7 ^{2}} = 1

What is 4x^2+16y^2-64=1 ?
The solution to 4x^2+16y^2-64=1 is Ellipse with (h,k)=(0,0),a=(sqrt(65))/2 ,b=(sqrt(65))/4