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Math Help - values-find word problem

  1. #1
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    values-find word problem

    1. find the value of k for which y= kx+1 is a tangent to the curve y^2 = 8x

    2. the hypotenuse of a triangle measures 17cm. The shortest side is 7cm shorter than the middle side. Find the lengths of the sides.

    really stuck. Have tried loads but my answers are wrong.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by lra11 View Post
    1. find the value of k for which y= kx+1 is a tangent to the curve y^2 = 8x
    .
    y=kx+1 is tangent to y^2=8x..
    therefore, they have a unique common point, say (a,b), that satisfies both equation.. so b=ka+1 \implies b^2 = k^2a^2 + 2ka + 1 and b^2=8a
    if we equate the b^2, we have k^2a^2 + 2ka + 1 = 8a \implies k^2a^2 + 2ka - 8a + 1 = k^2a^2 + a(2k - 8) + 1 = 0

    this is a quadratic equation hence, we can solve for the value of a:
    a = \frac{8-2k \pm \sqrt{(2k-8)^2 - 4k^2}}{2k^2}

    we want a unique solution for a, so we set the discriminant be equal to 0..

    (2k-8)^2 - 4k^2 = 0.. solve for k..

    Quote Originally Posted by lra11 View Post
    2. the hypotenuse of a triangle measures 17cm. The shortest side is 7cm shorter than the middle side. Find the lengths of the sides.

    really stuck. Have tried loads but my answers are wrong.
    if a is the lenth of the longer side, then a-7 is the length of the shorter side,

    since "hypotenuse" came, it can be assumed that this is a right triangle, hence 17^2 = a^2 + (a-7)^2.. do the algebra..
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