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

The solution to 4x^2+7y^2<= 2 is Ellipse with (h,k)=(0,0),a=sqrt(1/2),b=sqrt(2/7)

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4x^2+7y^2<= 2

4 x^{2} + 7 y^{2} \leq 2

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

Solution steps

4 x^{2} + 7 y^{2} \leq 2

Rewrite

4 x^{2} + 7 y^{2} \leq 2

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

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

\frac{x ^{2} - 4}{6 ^{2}} + \frac{y ^{2} + 4}{2 ^{2}} = 1

4 (x - 2)^{2} + (y - 3)^{2} - 1 = 0

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

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

What is 4x^2+7y^2<= 2 ?
The solution to 4x^2+7y^2<= 2 is Ellipse with (h,k)=(0,0),a=sqrt(1/2),b=sqrt(2/7)