What is 0.5=(2x^2-y^2)/(10) ?

The solution to 0.5=(2x^2-y^2)/(10) is Hyperbola with (h,k)=(0,0),a=sqrt(5/2),b=sqrt(5)

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0.5=(2x^2-y^2)/(10)

0.5 = \frac{2 x ^{2} - y ^{2}}{10}

(h, k) = (0, 0), a = \frac{5}{2}, b = 5

Solution steps

0.5 = \frac{2 x ^{2} - y ^{2}}{10}

Rewrite

0.5 - \frac{2 x ^{2} - y ^{2}}{10} = 0

in the form of a standard hyperbola equation

\frac{( x - 0 ) ^{2}}{( \frac{5}{2} ) ^{2}} - \frac{( y - 0 ) ^{2}}{( 5 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (0, 0), a = \frac{5}{2}, b = 5

5 x^{2} - 9 y^{2} + 10 x - 36 y - 14 = 0

x^{2} - y^{2} + 2 x + 5 = 0

x^{2} - y^{2} - 2 x - 8 y - 10 = 0

y^{2} = x^{2} + 361

4 x^{2} - y^{2} + 4 = 0

What is 0.5=(2x^2-y^2)/(10) ?
The solution to 0.5=(2x^2-y^2)/(10) is Hyperbola with (h,k)=(0,0),a=sqrt(5/2),b=sqrt(5)