Indentify the equation 9(x+1)^2=36+4y^2

Question

Solve the problem

Answer 1

This looks like the equation of a conic section.

9(x+1)²-4y²=36 is the equation of a hyperbola, centre (-1,0).

This can be standardised: (x+1)²/2²-y²/3²=1, where a=2 and b=3 in (x+1)²/a²-y²/b²=1. c²=a²+b²=13, making c (focal length)=√13.

The hyperbola cuts the x-axis when y=0, that is, when (x+1)²/4=1, (x+1)/2=±1, x+1=±2, making x=1 or -3.

The foci lie on the x-axis at (-1-√13,0) and (-1+√13,0), that is, √13 on either side of the centre (-1,0).

The asymptotes are found by solving (x+1)²/2²-y²/3²=0=((x+1)/2+y/3)((x+1)/2-y/3)=0,

This produces two linear graphs: 3(x+1)+2y=0 and 3(x+1)-2y=0, that is: y=-3(x+1)/2 and y=3(x+1)/2. These straight lines enclose the hyperbola.