Question

In problems$11-18$, find an equation of the conic described. Graph the equation by hand.

Vertices at $(0,1)$and $(6,1)$; asymptote the line$3y+2x=9$

Short answer

The equation of the hyperbola is$\frac{{(x-3)}^2}{9}-\frac{{(y-1)}^2}{4}=1$

Step-by-step solution

Step 1. Given Information

Co-ordinate of vertices $(0,1)\&(6,1)$

The given asymptote is$3y+2x=9$

Step 2. Calculation

Since $(a+\alpha ,\beta )=(0,1)$

and $(a-\alpha ,\beta )=(6,1)$

comparing the above two we get,

$\alpha =3,\beta =1,a=-3$

Step 3. Calculation 

The asymptote is $y=-\frac{2}{3}x+3$

Then we get, $a=-3,b=-2$

then the equation is$\frac{{(x-3)}^2}{9}-\frac{{(y-1)}^2}{4}=1$

Step 4. Graph