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samnatha Group Title

find the value of K for which x + 4y + K = 0 is a tangent to x^2 + y^2 - 2x + 2y - 15 = 0

  • one year ago
  • one year ago

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  1. campbell_st Group Title
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    you need to find the point(s) of contact before you can find K you could solve simultaneously and finding the 1 point of contact between the curve and the tangent. Substitution seems the most obvious method. An alternative method is to differentiate and let the derivative equal the slope of the tangent (-1/4) then solve for x.. once you know x find y by substituting. When you have the point of contact substitute into the tangent equation and solve for K

    • one year ago
  2. samnatha Group Title
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    i tried your first method and subed in -4y - k for x i then got a big long equation that went like this 17y^2 + K^2 + 8yk + 2K + 10 y = 0

    • one year ago
  3. campbell_st Group Title
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    ok because you are dealing with a circle then there are 2 points where the the tangents can occur |dw:1354733396218:dw| so rewrite theequation as 2 parts \[y = \sqrt{17 - (x -1)^2} - 1... and ....y = - \sqrt{17 - (x-1)^2} -1\] differentiate both of those.... let the derivatives equal -1/4 and solve for x.

    • one year ago
  4. samnatha Group Title
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    we have not done differentiate yet so i have no idea how to do it that way

    • one year ago
  5. campbell_st Group Title
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    ok.... that makes it difficult..

    • one year ago
  6. samnatha Group Title
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    yeah it does how would i do it the other way do u know ?

    • one year ago
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