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

Jul 18, 2016

A hyperbole $9 {\left(x - 2\right)}^{2} - {\left(y - 2\right)}^{2} - 9 = 0$ with asymptotes
$\left(x - 2\right) + \frac{1}{3} \left(y - 2\right) = 0$ and
$\left(x - 2\right) - \frac{1}{3} \left(y - 2\right) = 0$

#### Explanation:

This conic can be represented formally with the structure

$h \left(x , y\right) = {\left(x - {x}_{0}\right)}^{2} / a - {\left(y - {y}_{0}\right)}^{2} / b + c = 0$

and is characterized as a hyperbole.

Forcing

$9 {x}^{2} - {y}^{2} - 36 x + 4 y + 23 - h \left(x , y\right) = 0 , \forall \left\{x , y\right\} \in {\mathbb{R}}^{2}$

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

$\left\{a = \frac{1}{9} , b = 1 , c = - 9 , {x}_{0} = 2 , {y}_{0} = 2\right\}$

characterizing $h \left(x , y\right)$ as

$9 {\left(x - 2\right)}^{2} - {\left(y - 2\right)}^{2} - 9 = 0$

Also $h \left(x , y\right)$ can be written as

$\left(\sqrt{b} \left(x - {x}_{0}\right) + \sqrt{a} \left(y - {y}_{0}\right)\right) \left(\sqrt{b} \left(x - {x}_{0}\right) - \sqrt{a} \left(y - {y}_{0}\right)\right) + a b c = 0$

The two lines

$\sqrt{b} \left(x - {x}_{0}\right) + \sqrt{a} \left(y - {y}_{0}\right) = 0$ and
$\sqrt{b} \left(x - {x}_{0}\right) - \sqrt{a} \left(y - {y}_{0}\right) = 0$ are the asymptotes, or

$\left(x - 2\right) + \frac{1}{3} \left(y - 2\right) = 0$ and
$\left(x - 2\right) - \frac{1}{3} \left(y - 2\right) = 0$

Jul 18, 2016

$\left(1\right)$ Vertices : $\left(3 , 2\right) , \left(1 , 2\right) , \left(2 , 5\right) , \left(2 , - 1\right)$

$\left(2\right)$ Focii : $\left(2 + \sqrt{10} , 2\right) , \left(2 - \sqrt{10} , 2\right)$

$\left(3\right)$ Eqns. of Asymptotes : $3 x - y - 4 = 0 , 3 x + y - 8 = 0$

#### Explanation:

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

$9 {x}^{2} - 36 x + 36 - {y}^{2} + 4 y - 4 = - 23 + 36 - 4$

$\therefore 9 \left({x}^{2} - 4 x + 4\right) - \left({y}^{2} - 4 y + 4\right) = 9$

$\therefore 9 {\left(x - 2\right)}^{2} - {\left(y - 2\right)}^{2} = 9$

$\therefore {\left(x - 2\right)}^{2} / 1 - {\left(y - 2\right)}^{2} / 9 = 1$

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 $\left(X , Y\right)$ system, are (+-a,0)=(+-1,0) and

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

In $\left(x , y\right)$, they are, $x - 2 = \pm 1 , y - 2 = 0$, i.e., $x = 2 \pm 1 , y = 2$, or, (3,2), &,

(1,2).

$\left(2 , 2 \pm 3\right) , i . e . , \left(2 , 5\right) \mathmr{and} \left(2 , - 1\right)$ are also vertices.

The Eccentricity $e$ is given by,

${b}^{2} = {a}^{2} \left({e}^{2} - 1\right) \Rightarrow 9 = {e}^{2} - 1 \Rightarrow {e}^{2} = 10 \Rightarrow e = \sqrt{10}$.

Focii , in $\left(X , Y\right)$ system, are $\left(\pm a e , 0\right) = \left(\pm \sqrt{10} , 0\right)$

$\therefore$, In $\left(x , y\right)$ system, Focii are given by, $x - 2 = \pm \sqrt{10} , y - 2 = 0$ i.e., 3x=2+-sqrt10, y=2#.

$\therefore$ Focii, in original system, are $\left(2 \pm \sqrt{10} , 2\right)$

Asymptotes in $\left(X , Y\right)$ are, ${X}^{2} / 1 - {Y}^{2} / 9 = 0$, i.e., $Y = \pm 3 X$.

So, in $\left(x , y\right)$, they are, $y - 2 = \pm 3 \left(x - 2\right)$ or, $3 x - y - 4 = 0 , 3 x + y - 8 = 0$

Altogether,

$\left(1\right)$ Vertices : $\left(3 , 2\right) , \left(1 , 2\right) , \left(2 , 5\right) , \left(2 , - 1\right)$

$\left(2\right)$ Focii : $\left(2 + \sqrt{10} , 2\right) , \left(2 - \sqrt{10} , 2\right)$

$\left(3\right)$ Eqns. of Asymptotes : $3 x - y - 4 = 0 , 3 x + y - 8 = 0$