What is 4x^2+7y^2<28 ?

The solution to 4x^2+7y^2<28 is Ellipse with (h,k)=(0,0),a=sqrt(7),b=2

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4x^2+7y^2<28

4 x^{2} + 7 y^{2} < 28

(h, k) = (0, 0), a = 7, b = 2

Solution steps

4 x^{2} + 7 y^{2} < 28

Rewrite

4 x^{2} + 7 y^{2} < 28

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 0), a = 7, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 0), a = 7, b = 2

x^{2} + 9 y^{2} - 8 x + 36 y - 2 = 0

3 x^{2} + 9 x + y^{2} = 36 - (2 x^{2} + y^{2} - 3 x)

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} - 1 \geq 0

\frac{( x + 3 ) ^{2}}{3 ^{2}} + \frac{( y - 6 ) ^{2}}{6 ^{2}} = 1

\frac{x ^{2}}{4} + \frac{y ^{2}}{84} = 1

What is 4x^2+7y^2<28 ?
The solution to 4x^2+7y^2<28 is Ellipse with (h,k)=(0,0),a=sqrt(7),b=2