Question

How do you find all the critical points to graph `9x^2 – y^2 – 36x + 4y + 23 = 0` including vertices, foci and asymptotes?

Answer 1

A hyperbole `9(x-2)^2-(y-2)^2-9=0` with asymptotes
`(x-2)+1/3(y-2)=0` and
`(x-2)-1/3(y-2)=0`

Explanation:

This conic can be represented formally with the structure

`h(x,y)=(x-x_0)^2/a-(y-y_0)^2/b+c=0`

and is characterized as a hyperbole.

Forcing

`9 x^2 - y^2 - 36 x + 4 y + 23-h(x,y)=0, forall {x,y} in RR^2`

we have the conditions

#{ (c + x_0^2/a - y_0^2/b -23= 0), ( (2 y_0)/b-4 = 0), ( 1 - 1/b = 0), (36 - (2 x_0)/a = 0), (1/a -9= 0) :}#

Solving for `a,b,c,x_0,y_0` we obtain

`{a = 1/9, b = 1, c = -9, x_0 = 2, y_0 = 2}`

characterizing `h(x,y)` as

`9(x-2)^2-(y-2)^2-9=0`

Also `h(x,y)` can be written as

`(sqrt(b)(x-x_0)+sqrt(a)(y-y_0))(sqrt(b)(x-x_0)-sqrt(a)(y-y_0))+a b c=0`

The two lines

`sqrt(b)(x-x_0)+sqrt(a)(y-y_0)=0` and
`sqrt(b)(x-x_0)-sqrt(a)(y-y_0)=0` are the asymptotes, or

`(x-2)+1/3(y-2)=0` and
`(x-2)-1/3(y-2)=0`

Answer 2

`(1)` Vertices : `(3,2), (1,2), (2,5), (2,-1)`

`(2)` Focii : `(2+sqrt10,2), (2-sqrt10,2)`

`(3)` Eqns. of Asymptotes : `3x-y-4=0, 3x+y-8=0`

Explanation:

We complete the squares in the given eqn. to get the modified eqn.

`9x^2-36x+36-y^2+4y-4=-23+36-4`

`:.9(x^2-4x+4)-(y^2-4y+4)=9`

`:.9(x-2)^2-(y-2)^2=9`

`:. (x-2)^2/1-(y-2)^2/9=1`

Letting, `x-2=X, y-2=Y`, we have,

`X^2/1-Y^2/9=1`, which represents a Hyperbola .

Comparing with, `X^2/a^2-Y^2/b^2=1`, we have,

`a^2=1,b^2=9`

Hence the Vertices, in `(X,Y)` system, are #(+-a,0)=(+-1,0) and

(0,+-b)=(0,+-3)#

In `(x,y)`, they are, `x-2=+-1,y-2=0`, i.e., `x=2+-1,y=2`, or, #(3,2), &,

(1,2)#.

`(2,2+-3), i.e., (2,5) and (2,-1)` are also vertices.

The Eccentricity `e` is given by,

`b^2=a^2(e^2-1) rArr 9=e^2-1rArre^2=10rArre=sqrt10`.

Focii , in `(X,Y)` system, are `(+-ae,0)=(+-sqrt10,0)`

`:.`, In `(x,y)` system, Focii are given by, `x-2=+-sqrt10, y-2=0` i.e., 3x=2+-sqrt10, y=2#.

`:.` Focii, in original system, are `(2+-sqrt10,2)`

Asymptotes in `(X,Y)` are, `X^2/1-Y^2/9=0`, i.e., `Y=+-3X`.

So, in `(x,y)`, they are, `y-2=+-3(x-2)` or, `3x-y-4=0, 3x+y-8=0`

Altogether,

`(1)` Vertices : `(3,2), (1,2), (2,5), (2,-1)`

`(2)` Focii : `(2+sqrt10,2), (2-sqrt10,2)`

`(3)` Eqns. of Asymptotes : `3x-y-4=0, 3x+y-8=0`