What is x^2+(y^2)/4 <1 ?

The solution to x^2+(y^2)/4 <1 is Ellipse with (h,k)=(0,0),a=1,b=2

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x^2+(y^2)/4 <1

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

(h, k) = (0, 0), a = 1, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 1, b = 2

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (0, 0), b = 2, a = 1

16 x^{2} + 25 y^{2} = 1

4 x^{2} + 2 y^{2} - 8 x + 12 y + 6 = 0

\frac{( x + 3 ) ^{2}}{25} + \frac{( y - 8 ) ^{2}}{16} = 1

16 x^{2} + 9 y^{2} - 36 y = 108

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

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