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Math Help - Parabolas

  1. #1
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    Exclamation Parabolas

    Hi

    Write down the cartesian equation of the tangent + normal to the following parabolas at the points indicated.

    (a) x = cos 2t, y = cos t at the point where t = PI/4

    (b) x = t + 2, y = t^2/4 + 1 at the vertex.

    For (a), I keep on getting 2x - 4sqrt(2) + 4 = 0 and 4x + sqrt(2)y -1 = 0 for the tangent and the normal respectively, which is wrong.

    For (b), I have no clue how to do it.

    Could someone please help me?
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by xwrathbringerx View Post
    For (a), I keep on getting 2x - 4sqrt(2)y + 4 = 0 and 4x + sqrt(2)y -1 = 0 for the tangent and the normal respectively, which is wrong.
    I also the same equations as you did. Why are they wrong?
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  3. #3
    AMI
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    Your answer was correct (except for the missing "y"). Where did you get stuck at (b)? Finding the vertex?!
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  4. #4
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    Exclamation

    Thanx - that must mean the textbook answer's wrong

    Yeah, for (b), I have no clue how to find the vertex.
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  5. #5
    AMI
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    y=\frac{1}{4}t^2+1,x-2=t \Rightarrow y=\frac{1}{4}(x-2)^2+1 and you can see that this parabola is in the vertex form: y=a(x-x_v)^2+y_v
    (\Rightarrow y\geqslant y_v,\forall x\text{ and }\exists x=x_v\text{ so that }y=y_v\Rightarrow (x_v,y_v)\text{ is the vertex})
    So you have (x_v,y_v)=(2,1).
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