Question

How do you find all the critical points to graph `36x^2 - 100y^2 - 72x + 400y = 3964` including vertices, foci and asymptotes?

Answer 1

Vertices are `(11,2)` and `(-9,2)`, foci are (`1±2sqrt(34),2`) and asymptotes are `y=3/5x+7/5` and `y= -3/5x+13/5`.

Explanation:

The graph is for a hyperbola.
The standard form of equation for a hyperbola is
`x^2/(a^2) - y^2/(b^2) = ±1`. `(a>0, b>0)`
What we have to do is to deform the fomula to this form.

[step1] Let's complete the square.
`36x^2 -100y^2 -72x +400y =3964`
`36(x^2-2x) -100(y^2-4y)=3964`
`36(x-1)^2-100(y-2)^2=3964+36×1-100×4`
`36(x-1)^2-100(y-2)^2=3600`

[step2] divide both side by `3600` to make a `1` on the right side.
`(x-1)^2/(10^2) - (y-2)^2/(6^2) = 1`

The hyperbola has vertices and foci on on the line `y=2`.
Substituting `y=2` to the fomula yields `x=11,-9`
Therefore, the vertices are `(11,2)` and `(-9,2)`.

The foci of `x^2/(a^2) - y^2/(b^2) = 1` are `(±sqrt(a^2+b^2),0)` .
If the fomula were `x^2/(10^2) - y^2/(6^2) = 1`, the foci would be
(`±sqrt(6^2+10^2),0`) = (`±2sqrt(34),0`). Translate this `1` to the right and `2` to the up and the foci are (`1±2sqrt(34),2`).

The asymptopes of `x^2/(a^2) - y^2/(b^2) = 1` are `y=±(a/b)x` and the asymptopes of `(x-1)^2/(10^2) - (y-2)^2/(6^2) = 1` are therefore
`y-2 = ±(6/10)(x-1)`
`y=3/5x+7/5` and `y= -3/5x+13/5`.