What is y^2+1=x^2+8x ?

The solution to y^2+1=x^2+8x is Hyperbola with (h,k)=(-4,0),a=sqrt(17),b=sqrt(17)

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y^2+1=x^2+8x

y^{2} + 1 = x^{2} + 8 x

(h, k) = (- 4, 0), a = 17, b = 17

Solution steps

y^{2} + 1 = x^{2} + 8 x

Rewrite

y^{2} + 1 = x^{2} + 8 x

in the form of a standard hyperbola equation

\frac{( x - ( - 4 ) ) ^{2}}{( 17 ) ^{2}} - \frac{( y - 0 ) ^{2}}{( 17 ) ^{2}} = 1

Therefore Hyperbola properties are:

(h, k) = (- 4, 0), a = 17, b = 17

9 x^{2} - y^{2} = 1

5 x^{2} - 5 y^{2} + 20 x - 60 y = 185

\frac{x ^{2}}{9} - \frac{y ^{2}}{1} = 1

x^{2} - 5 y^{2} = 3 y

\frac{y ^{2}}{81} - \frac{x ^{2}}{25} = 1

What is y^2+1=x^2+8x ?
The solution to y^2+1=x^2+8x is Hyperbola with (h,k)=(-4,0),a=sqrt(17),b=sqrt(17)