Question

24) Find center vertices foci asymptotes

Answer 1

The given equation represents a hyperbola. Its main features are:

1. center - (1,-3)
2. vertices - V1=  (1,-3+$\sqrt{2}$) / V2= (1,-3-$\sqrt{2}$)
3. foci - F1=  (1 , -3+ $2\sqrt{5}$ )/ F2= (1 , -3 - $2\sqrt{5}$ )
4. asymptotes = $\pm \frac{1}{3}\left(x-1\right)-3$

Hyperbola

A hyperbola can be defined by its center, vertices,  foci and asymptotes. And it is represented algebraically  by the standard equation:  $\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$, where:

h= x-coordinate of center

k= y-coordinate of center

a and b= semi-axis

First, you need to rewrite the given equation 9y²-x²+2x+54y+62=0 in the standard equation hyperbola:

$(-x^2+2x+?)+9(y^2+6y+?)=-62\\(-x^2+2x+?)+9(y^2+6y+9)=-62\\(-x^2+2x-1)+9(y^2+6y+9)=-62\\(-x^2+2x-1)+9(y^2+6y+9)=-62-1+81\\((-x^2+2x-1)+9(y^2+6y+9)=18 ) \div 18\\ \frac{(-x^2+2x-1)}{18} +\frac{9(y^2+6y+9}{18}= \frac{18}{18} \\\frac{-(x-1)^2}{18} +\frac{(y+3^2)}{2}=1\\\frac{-(x-1)^2}{(3\sqrt{2})^2} +\frac{(y+3^2)}{(\sqrt{2})^2 }=1\\\frac{\left(y+3)^2}{\left(\sqrt{2}\right)^2}-\frac{\left(x-1\right)^2}{\left(3\sqrt{2}\right)^2}=1$

Comparing the previous equation with the standard form ($\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1$), you have:

h=1,  k=-3,  a=$\sqrt{2}$ and b=$3\sqrt{2}$ . From now, it is possible to find that the question asks:

• Find the center

The coordinates for center is (h,k). Thus, the center is (1,-3).

• Find the vertices

The vertices (V1 and V2) of hyperbola can be found from the coordinates of center (h,k) and the semi-axis (a).

V1= (h,k+a)= (1,-3+$\sqrt{2}$)

V2= (h,k-a)= (1,-3-$\sqrt{2}$)

• Find the foci

The foci or the focus points can be found from the coordinates of center (h,k) and the c ($\sqrt{a^2+b^2}$) which represents the distance from the center to the focus.

$c=\sqrt{a^2+b^2}\\ c=\sqrt{(3\sqrt{2} )^2+(\sqrt{2} )^2}\\c=\sqrt{18+2} \\c=\sqrt{20}=2\sqrt{5}$

Thus,

F1= (h,k+c)= (1 , -3+ $2\sqrt{5}$ )

F2= (h,k-c)= (1 , -3 - $2\sqrt{5}$ )

• Find the asymptotes

The asymptotes are the lines the hyperbola tends to at ±∞. For hyperbola, the asymptotes are defined as: $y=\pm \frac{a}{b}\left(x-h\right)+k$. Then, for this question:

$y=\pm \frac{\sqrt{2}}{3\sqrt{2}}\left(x-1\right)-3\\y=\pm \frac{1}{3}\left(x-1\right)-3$

Read more about hyperbola here:

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