Answer:
Step-by-step explanation:
given that the pair of parametric equations represents a line, parabola, circle, ellipse, or hyperbola.
We are given
[tex]x=3cosh (4t) \\y = 4sinh(4t)\\[/tex]
We know in hyperbolic functions
[tex]cosh^2 (4t) -sinh^2 (4t) =1[/tex]
Using this we can find out
[tex]cosh (4t) = \frac{x}{3} \\sinh(4t)=\frac{y}{4}[/tex]
Using the identity we can say that
[tex]\frac{x^2}{9} -\frac{y^2}{16} =1[/tex]
this is nothing but a hyperbola.
a =3 and b =4