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samnatha

  • 2 years ago

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

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  1. campbell_st
    • 2 years ago
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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

  2. samnatha
    • 2 years ago
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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

  3. campbell_st
    • 2 years ago
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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.

  4. samnatha
    • 2 years ago
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    we have not done differentiate yet so i have no idea how to do it that way

  5. campbell_st
    • 2 years ago
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    ok.... that makes it difficult..

  6. samnatha
    • 2 years ago
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    yeah it does how would i do it the other way do u know ?

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