Answer :
Given:
An architect has designed two tunnels.
Tunnel A is modeled by:
[tex]x^2+y^2+30x+56=0[/tex]Tunnel B is modeled by:
[tex]x^2-30x+16y-95=0[/tex]Part A: write the equation of tunnel A in standard from
So, we will complete the square for the two variables x, and y as follows:
[tex]\begin{gathered} (x^2+30x)+y^2=-56 \\ (x^2+30x+225)+y^2=-56+225 \\ (x+15)^2+y^2=169 \\ (x+15)^2+y^2=13^2 \end{gathered}[/tex]So, the answer to part A:
The equation will be: (x+15)² + y² = 13²
The conic section is: Circle with canter (-15, 0) and radius = 13 feet
Part B: Write the equation of tunnel B in standard from
So, we will complete the square for the variable x as follows:
[tex]\begin{gathered} (x^2-30x)=-16y+95 \\ (x^2-30x+225)=-16y+95+225 \\ (x-15)^2=-16y+320 \\ (x-15)^2=-16(y-20) \\ (x-15)^2=-4(4)(y-20) \end{gathered}[/tex]So, the answer to part B:
The equation will be: (x-15)² = -4(4)(y-20)
The conic section is: a Parabole with the vertex at (15, 20)
Part C: Determine the maximum height of each tunnel.
Tunnel A:
[tex](x+15)^2+y^2=13^2[/tex]So, the radius = 13, so, the diameter = 2 * 13 = 26
So, the maximum height = 13 feet
The truck has a height = of 13.5 feet
so, it will not be able to pass through the tunnel A
Tunnel B:
[tex](x-15)^2=-16(y-20)[/tex]when y = 0, solve for x:
[tex]\begin{gathered} (x-15)^2=-16*-20 \\ (x-15)^2=320 \\ x-15=\pm\sqrt{320} \\ x=15\pm\sqrt{320}=32.88,-2.88 \end{gathered}[/tex]When x = 15, we will find the maximum height
[tex]\begin{gathered} 0=-16(y-20) \\ y-20=0 \\ y=20 \end{gathered}[/tex]So, the maximum height = 20 feet
the truck will be able to pass through the tunnel