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

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- Feb 26th 2011, 03:05 AMPunchtwo distinct points
Given the equation $\displaystyle \frac{(x-3)^2}{4}-y^2=1$ and $\displaystyle 16x^2-64x+64+a^2y^2-16a^2=0$, determine the range of value of $\displaystyle a$ such that the two graphs intersect at exactly two distinct points.

- Feb 26th 2011, 05:08 AMSoroban
Hello, Punch!

Quote:

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

$\displaystyle \text{determine the range of value of }a\text{ such that the two graphs}$

. . $\displaystyle \text{intersect at exactly two distinct points.}$

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

$\displaystyle \begin{array}{ccccccc}

\text{Divide [1] by 16:} & \dfrac{(x-3)^2}{64} - \dfrac{y^2}{16} &=& \dfrac{1}{16} \\ \\[-3mm]

\text{Subtract [3]:} & \text{-}\dfrac{(x-2)^2}{a^2} + \dfrac{y^2}{16} &=& \text{-}1 \end{array}$

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

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

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

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

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

. . $\displaystyle Go\;for\;it!$

- Feb 26th 2011, 12:58 PMWilmer
- Feb 26th 2011, 05:59 PMPunch

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

http://i952.photobucket.com/albums/a...g/P2240112.jpg - Feb 26th 2011, 06:18 PMWilmer
Your 272a^4 + 17152a^2 simplifies to 16a^2(17a^2 + 1072),

which is same as what I showed you in my previous post. - Feb 26th 2011, 06:41 PMPunch
- Feb 26th 2011, 06:47 PMPunch
Also, is it possible to show how you could transform $\displaystyle \frac{(x-3)^2}{4}-y^2=1$ into $\displaystyle \;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;$?

- Feb 26th 2011, 08:22 PMWilmer
- Feb 26th 2011, 08:46 PMPunch