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    Line Y = 3X + 1
    Curve X^2 + Y^2 = K

    FIND VALUES OF K FOR WHICH THE LINE IS A TANGENT TO THE CURVE?

    IM DOING WJEC SO PLEASE USE A METHOD WHICH IS RELEVANT

    This is how I tried it:

    As Y (line) = Y (curve) therefore
    (3X + 1)^2 = K - X^2
    I used the fact that Y = Y

    So we get

    (3X +1)(3X+1)= 9X^2 + 3X + 3X +1
    9X^2 + 3X + 3X +1 = K - X^2
    8X ^2 + 6X + 1 - K = 0

    So now we can use the discriminant on it, as its in ax^2 + bx + c = 0 form.

    a = 8
    b = 6
    c = 1 - K (not sure about this)

    B^2 - 4AC
    (6)^2 - 4(8)(K+1) = 0
    36 - 32(K+1) = 0
    36 - 32K - 32 = 0
    4 = 32K
    4/32 = K
    1/8 = K

    Im I right help please!
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    You've got the right method, but you made an error here:
    (Original post by Tempa)
    9X^2 + 3X + 3X +1 = K - X^2
    8X ^2 + 6X + 1 - K = 0
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    9X^2 + 3X + 3X +1 = K - X^2

    This ^ is correct but the line after it is wrong. Can you correct it?
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    as far as i know 9x^2 - x^2 is 8x^2?
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    sh!!!t
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    (Original post by Tempa)
    as far as i know 9x^2 - x^2 is 8x^2?
    But the -x² is on the other side of the equation, so you have to add it.

    Silly mistake.
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    9X^2 + 3X + 3X +1 = K - X^2
    10 ^2 + 6X + 1 - K = 0

    So now we can use the discriminant on it, as its in ax^2 + bx + c = 0 form.

    a = 10
    b = 6
    c = 1 - K (not sure about this)

    B^2 - 4AC
    (6)^2 - 4(10)(K+1) = 0
    36 - 40(K+1) = 0
    36 - 40k - 40 = 0
    -4 - 40k = 0
    40k= -4
    k = -0.1 or -1/10

    is this correct?
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    (Original post by Get me off the £\?%!^@ computer)
    Sarcasm? People are trying to help you.
    nah not sarcasm mate just i make silly mistakes like that all the time its annoying
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    (Original post by Tempa)
    9X^2 + 3X + 3X +1 = K - X^2
    10 ^2 + 6X + 1 - K = 0

    So now we can use the discriminant on it, as its in ax^2 + bx + c = 0 form.

    a = 10
    b = 6
    c = 1 - K (not sure about this)

    B^2 - 4AC
    (6)^2 - 4(10)(K+1) = 0
    36 - 40(K+1) = 0
    36 - 40k - 40 = 0
    -4 - 40k = 0
    40k= -4
    k = -0.1 or -1/10

    is this correct?
    Nope, your 1-k magically became k+1 somehow (it should be 1-k).
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    ah sh1t another one of my silly mistakes fml
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    9X^2 + 3X + 3X +1 = K - X^2
    10 ^2 + 6X + 1 - K = 0

    So now we can use the discriminant on it, as its in ax^2 + bx + c = 0 form.

    a = 10
    b = 6
    c = 1 - K (not sure about this)

    B^2 - 4AC
    (6)^2 - 4(10)(K-1) = 0
    36 - 40(K-1) = 0
    36 - 40k + 40 = 0
    76 - 40k = 0
    76 = 40k
    k = 76/40
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    9X^2 + 3X + 3X +1 = K - X^2
    10 ^2 + 6X + 1 - K = 0

    So now we can use the discriminant on it, as its in ax^2 + bx + c = 0 form.

    a = 10
    b = 6
    c = 1 - K (not sure about this)

    B^2 - 4AC
    (6)^2 - 4(10)(K-1) = 0
    36 - 40(K-1) = 0
    36 - 40k + 40 = 0
    76 - 40k = 0
    76 = 40k
    k = 76/40
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    accident, but im i correct?
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    (Original post by Tempa)
    accident, but im i correct?
    Nope, your 1-k magically became k-1. It should still be 1-k.
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    so is k = 1/10
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    (Original post by nuodai)
    Nope, your 1-k magically became k-1. It should still be 1-k.
    1/10?
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    (Original post by Tempa)
    1/10?
    Got there eventually! Well done.
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    (Original post by nuodai)
    Got there eventually! Well done.
    any tips to improve my maths i need a A* in alev

    had a flukey a* at gcse
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    (Original post by Tempa)
    any tips to improve my maths i need a A* in alev

    had a flukey a* at gcse
    Just do as many practice questions as you can really. Also try checking your answers for yourself without asking other people -- it'll help you get better at spotting common and silly mistakes (like the ones you were making here). Like here you could have put K=\dfrac{1}{8} into the second equation, substituted the first equation into the second equation and used the discriminant to make sure there is only one root. [That's what I did to check your answer.]
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    (Original post by Tempa)
    any tips to improve my maths i need a A* in alev

    had a flukey a* at gcse
    I got by 2ums at GCSE, and 1ums at a2. Soloman past papers all the way.
 
 
 
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