Answer:
Tunnel A: Circle
Tunnel B: Parabola
Max height of A: 12 ft
Max height of B: 16 ft
The truck can only pass through tunnel B.
Explanation:
Since we do not know what x and y represent, we assume that is the height of the tunnel and x is the width.
Part A:
Let us convert our equation into the standard form.
The equation for tunnel A is
[tex]x^2+y^2+28x+52=0[/tex]which we rewrite as
[tex](x^2+28x+\cdots)+y^2=-52[/tex]Now we complete the square for variable x. What should we add to x^2 + 28x to make it a complete square?
After some thinking, we realise that we do x^2 + 28x + 14^2 then we have (x + 14)^2 .
Therefore, we add 14^2 to both sides of our equation to get:
[tex](x^2+28x+14^2)+y^2=-52+14^2[/tex][tex](x+14)^2+y^2=-52+14^2[/tex][tex](x+14)^2+y^2=144[/tex]this equation we recognise as that of a circle! Therefore, the conic section for tunnel A is a circle.
Part B:
Let us now turn to tunnel B and write its equation:
[tex]x^2-36x+16y+68=0[/tex]The first thing to note is that the above equation is linear in y; therefore, we can rearrange the equation to write it as
[tex]16y=-(x^2-36x+68)[/tex]Now we have to complete the square on the right-hand side.
subtracting 256 from both sides gives
[tex]16y-256=-(x^2-36x+68)-256[/tex][tex]16y-256=-x^2+36x-324[/tex][tex]16y-256=-(x^2-36x+324)[/tex][tex]16y-256=-(x-18)^2[/tex][tex]\Rightarrow y=-\frac{1}{16}(x-18)^2+\frac{256}{16}[/tex]which is the standard equation for a parabola!
Hence, the conic section for tunnel B is that of a parabola.
