Asked by Anonymous
                how do you sketch the curve represented by 
x=(√t2+2)
y=t/4?
            
            
        x=(√t2+2)
y=t/4?
Answers
                    Answered by
            Steve
            
    just like the other one!
pick some values for t, calculate x and y, and plot the points.
First, decide whether you meant
x=?(t^2+2)
or
x=?t^2 + 2)
I suspect the former, since ?t^2 = |t|
So, x is basically just t
y is basically just t.
The graph will be almost a straight line as x gets large. To wit:
http://www.wolframalpha.com/input/?i=parametric+plot+x%3D%E2%88%9A(t%5E2%2B2),+y%3Dt%2F4
Looks like a branch of an hyperbola, no? Let's see
t = 4y
x=?(t^2+2)
=?(y^2/16+2)
x^2 = y^2/16 + 2
x^2/2 - y^2/32 = 1
Yep - an hyperbola. For large x, the graph approaches the asymptotes.
    
pick some values for t, calculate x and y, and plot the points.
First, decide whether you meant
x=?(t^2+2)
or
x=?t^2 + 2)
I suspect the former, since ?t^2 = |t|
So, x is basically just t
y is basically just t.
The graph will be almost a straight line as x gets large. To wit:
http://www.wolframalpha.com/input/?i=parametric+plot+x%3D%E2%88%9A(t%5E2%2B2),+y%3Dt%2F4
Looks like a branch of an hyperbola, no? Let's see
t = 4y
x=?(t^2+2)
=?(y^2/16+2)
x^2 = y^2/16 + 2
x^2/2 - y^2/32 = 1
Yep - an hyperbola. For large x, the graph approaches the asymptotes.
                    Answered by
            Steve
            
    you can either
(a) find and fix my typo
(b) read bobpursley's solution, which is correct.
    
(a) find and fix my typo
(b) read bobpursley's solution, which is correct.
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