Question

Which of the following is an asymptote of the graph? (The points shown in blue are foci.)y−3=4/3xy minus 3 is equal to 4 thirds xy=5/3xy is equal to 5 thirds xy+3=−3/4xy plus 3 is equal to negative 3 fourths xy−3=−3/4x

Answer 1

The equations of the hyperbola are

`y=\pm(b)/(a)(x-h)+k`

(h, k) are the center of it

(h + a, k), (h - a, k) are the vertices

(h + c, k), (h - c, k) are the foci

From the graph, we can see the vertices are (-3, 3) and (3, 3), then

`\begin{gathered} h+a=3\rightarrow(1) \\ k=3 \end{gathered}`

The foci are (-5, 3) and (5, 3), then

`h+c=5\rightarrow(2)`

From the graph, we can see the center is the midpoint of the line joining the foci, then the center is (0, 3), then

h = 0

k = 3

Substitute the value of h in (1) and (2) to find a and c

`\begin{gathered} 0+a=3 \\ a=3 \\ 0+c=5 \\ c=5 \end{gathered}`

Use the relation

`\begin{gathered} c^2=a^2+b^2 \\ b^2=c^2-a^2 \\ b^2=5^2-3^2 \\ b^2=25-9=16 \\ b=\pm4 \end{gathered}`

Now, substitute the values of a, b, h, k in the equations of the asymptote above

`\begin{gathered} y=(4)/(3)(x-0)+3 \\ y=(4)/(3)x+3 \\ y-3=(4)/(3)x\rightarrow(1st) \end{gathered}`
`\begin{gathered} y=-(4)/(3)x+3 \\ y-3=-(4)/(3)x \end{gathered}`

The correct answer is A (1st equation of asymptote)