What is e^{ln(y^2+2)}=e^{ln(x^2+4)} ?

The solution to e^{ln(y^2+2)}=e^{ln(x^2+4)} is Hyperbola with (h,k)=(0,0),a=sqrt(2),b=sqrt(2)

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e^{ln(y^2+2)}=e^{ln(x^2+4)}

e^{l n (y^{2} + 2)} = e^{l n (x^{2} + 4)}

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

Solution steps

e^{l n (y^{2} + 2)} = e^{l n (x^{2} + 4)}

Rewrite

e^{l n (y^{2} + 2)} - e^{l n (x^{2} + 4)} = 0

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

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

144 x^{2} - 25 y^{2} = 169

3 x^{2} - 4 y^{2} + 16 y - 18 = 0

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

\frac{x ^{2}}{4} - \frac{y ^{2}}{21} = 1

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

What is e^{ln(y^2+2)}=e^{ln(x^2+4)} ?
The solution to e^{ln(y^2+2)}=e^{ln(x^2+4)} is Hyperbola with (h,k)=(0,0),a=sqrt(2),b=sqrt(2)