What is 10y^2+60y-160x-255=20x^2 ?

The solution to 10y^2+60y-160x-255=20x^2 is Hyperbola with (h,k)=(-4,-3),a=sqrt(5/2),b=(sqrt(5))/2

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10y^2+60y-160x-255=20x^2

10 y^{2} + 60 y - 160 x - 255 = 20 x^{2}

(h, k) = (- 4, - 3), a = \frac{5}{2}, b = \frac{5}{2}

Solution steps

10 y^{2} + 60 y - 160 x - 255 = 20 x^{2}

Rewrite

10 y^{2} + 60 y - 160 x - 255 = 20 x^{2}

in the form of a standard hyperbola equation

\frac{( y - ( - 3 ) ) ^{2}}{( \frac{5}{2} ) ^{2}} - \frac{( x - ( - 4 ) ) ^{2}}{( \frac{5}{2} ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 4, - 3), a = \frac{5}{2}, b = \frac{5}{2}

\frac{( y + 8 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

a^{2} - b^{2} = 3

a^{2} - b^{2} = 5

\frac{( y - 6 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{64} = 1

9 y^{2} - 4 x^{2} - 18 y + 24 x - 63 = 0

What is 10y^2+60y-160x-255=20x^2 ?
The solution to 10y^2+60y-160x-255=20x^2 is Hyperbola with (h,k)=(-4,-3),a=sqrt(5/2),b=(sqrt(5))/2