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Math Help - two distinct points

  1. #1
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    two distinct points

    Given the equation \frac{(x-3)^2}{4}-y^2=1 and 16x^2-64x+64+a^2y^2-16a^2=0, determine the range of value of a such that the two graphs intersect at exactly two distinct points.
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  2. #2
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    Hello, Punch!

    \text{Given the equations: }\:\begin{Bmatrix}[1] &\dfrac{(x-3)^2}{4}-y^2\:=\:1 \\ \\[-3mm] [2] & 16x^2-64x+64+a^2y^2-16a^2\:=\:0 \end{Bmatrix}

    \text{determine the range of value of }a\text{ such that the two graphs}
    . . \text{intersect at exactly two distinct points.}

    \text{Write [2] in standard form: }\;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;[3]


    \begin{array}{ccccccc}<br />
\text{Divide [1] by 16:} & \dfrac{(x-3)^2}{64} - \dfrac{y^2}{16} &=& \dfrac{1}{16} \\ \\[-3mm]<br />
\text{Subtract [3]:} & \text{-}\dfrac{(x-2)^2}{a^2} + \dfrac{y^2}{16} &=& \text{-}1 \end{array}

    \text{We have: }\quad\;\;\dfrac{(x-3)^2}{64} - \dfrac{(x-2)^2}{a^2} \;=\;\;\text{-}\dfrac{15}{16}


    \text{Multiply by }64a^2\!:\;\;a^2(x-3)^2 - 64(x-2)^2 \;=\;\text{-}60a^2


    \text{Simplify: }\;(a^2-64)x^2 + 250x + (69a^2 - 256) \;=\;0


    \text{Quadratic Formula: }\;x \;=\;\dfrac{-250 \pm\sqrt{250^2 - 4(a^2-64)(69a^2-256)}}{2(a^2-64)}


    \text{To have two intersections, the discriminant must be positive.}

    . . Go\;for\;it!

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  3. #3
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    Quote Originally Posted by Soroban View Post
    \text{Write [2] in standard form: }\;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;[3]
    Hmmm....Soroban, that should be +y^2 / 16 ; right?

    I end up with something quite different:
    x = [3a^2 + 128 +- 2aSQRT(17a^2 + 1072)] / (a^2 + 64).

    Think it's easier to substitute y^2 = (x^2 - 6x + 5) / 4 from [1] to [2].
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, Punch!


    \text{Write [2] in standard form: }\;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;[3]


    \begin{array}{ccccccc}<br />
\text{Divide [1] by 16:} & \dfrac{(x-3)^2}{64} - \dfrac{y^2}{16} &=& \dfrac{1}{16} \\ \\[-3mm]<br />
\text{Subtract [3]:} & \text{-}\dfrac{(x-2)^2}{a^2} + \dfrac{y^2}{16} &=& \text{-}1 \end{array}

    \text{We have: }\quad\;\;\dfrac{(x-3)^2}{64} - \dfrac{(x-2)^2}{a^2} \;=\;\;\text{-}\dfrac{15}{16}


    \text{Multiply by }64a^2\!:\;\;a^2(x-3)^2 - 64(x-2)^2 \;=\;\text{-}60a^2


    \text{Simplify: }\;(a^2-64)x^2 + 250x + (69a^2 - 256) \;=\;0


    \text{Quadratic Formula: }\;x \;=\;\dfrac{-250 \pm\sqrt{250^2 - 4(a^2-64)(69a^2-256)}}{2(a^2-64)}


    \text{To have two intersections, the discriminant must be positive.}

    . . Go\;for\;it!


    I tried another method which was to sub equation [1] into [2] and equating the discriminant, (b^2-4ac)>0 but it didn't work. Here's my working.

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  5. #5
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    Your 272a^4 + 17152a^2 simplifies to 16a^2(17a^2 + 1072),
    which is same as what I showed you in my previous post.
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  6. #6
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    Quote Originally Posted by Wilmer View Post
    Your 272a^4 + 17152a^2 simplifies to 16a^2(17a^2 + 1072),
    which is same as what I showed you in my previous post.
    okay then, maybe we should wait for soroban's reply
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  7. #7
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    Also, is it possible to show how you could transform \frac{(x-3)^2}{4}-y^2=1 into \;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;?
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  8. #8
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    Quote Originally Posted by Punch View Post
    Also, is it possible to show how you could transform \frac{(x-3)^2}{4}-y^2=1 into \;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;?
    That's not what Soroban did: he transformed the other equation.

    And made (I believe) a sign error; should be (x-2)^2 / a^2 + y^2 / 16 = 1

    Multiply that through by 16a^2 and you get:
    16x^2 - 64x + 64 +a^2y^2 - 16a^2 = 0
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  9. #9
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    Quote Originally Posted by Wilmer View Post
    That's not what Soroban did: he transformed the other equation.

    And made (I believe) a sign error; should be (x-2)^2 / a^2 + y^2 / 16 = 1

    Multiply that through by 16a^2 and you get:
    16x^2 - 64x + 64 +a^2y^2 - 16a^2 = 0
    Sorry! I see it now
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