What is (((x-3)^2)}{16}-\frac{((y+1)^2))/4 <1 ?

The solution to (((x-3)^2)}{16}-\frac{((y+1)^2))/4 <1 is Hyperbola with (h,k)=(3,-1),a=4,b=2

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(((x-3)^2)}{16}-\frac{((y+1)^2))/4 <1

\frac{( ( x - 3 ) ^{2} )}{16} - \frac{( ( y + 1 ) ^{2} )}{4} < 1

(h, k) = (3, - 1), a = 4, b = 2

Solution steps

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 1 ) ^{2}}{4} < 1

Rewrite

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 1 ) ^{2}}{4} < 1

in the form of a standard hyperbola equation

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

Therefore Hyperbola properties are:

(h, k) = (3, - 1), a = 4, b = 2

9 x^{2} - 4 y^{2} - 36 x - 16 x + 20 = 0

x^{2} - y^{2} + 3 x - 4 y = 6

4 x^{2} - 34 y - 5 y^{2} + 36 = 0

9 x^{2} - 4 y^{2} + 24 \leq 0

x^{2} - 16 y^{2} = 64

What is (((x-3)^2)}{16}-\frac{((y+1)^2))/4 <1 ?
The solution to (((x-3)^2)}{16}-\frac{((y+1)^2))/4 <1 is Hyperbola with (h,k)=(3,-1),a=4,b=2