# two distinct points

• Feb 26th 2011, 04:05 AM
Punch
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.
• Feb 26th 2011, 06:08 AM
Soroban
Hello, Punch!

Quote:

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

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

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

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

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

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

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

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

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

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

• Feb 26th 2011, 01:58 PM
Wilmer
Quote:

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

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].
• Feb 26th 2011, 06:59 PM
Punch
Quote:

Originally Posted by Soroban
Hello, Punch!

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

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

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

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

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

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

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

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

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.

http://i952.photobucket.com/albums/a...g/P2240112.jpg
• Feb 26th 2011, 07:18 PM
Wilmer
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, 07:41 PM
Punch
Quote:

Originally Posted by Wilmer
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
• Feb 26th 2011, 07:47 PM
Punch
Also, is it possible to show how you could transform $\frac{(x-3)^2}{4}-y^2=1$ into $\;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;$?
• Feb 26th 2011, 09:22 PM
Wilmer
Quote:

Originally Posted by Punch
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
• Feb 26th 2011, 09:46 PM
Punch
Quote:

Originally Posted by Wilmer
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