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.
Hello, Punch!
$\displaystyle \text{Given the equations: }\:\begin{Bmatrix}[1] &\dfrac{(x-3)^2}{4}-y^2\:=\:1 \\ \\[-3mm] [2] & 16x^2-64x+64+a^2y^2-16a^2\:=\:0 \end{Bmatrix}$
$\displaystyle \text{determine the range of value of }a\text{ such that the two graphs}$
. . $\displaystyle \text{intersect at exactly two distinct points.}$
$\displaystyle \text{Write [2] in standard form: }\;\dfrac{(x-2)^2}{a^2} - \dfrac{y^2}{16} \:=\:1\;\;[3]$
$\displaystyle \begin{array}{ccccccc}
\text{Divide [1] by 16:} & \dfrac{(x-3)^2}{64} - \dfrac{y^2}{16} &=& \dfrac{1}{16} \\ \\[-3mm]
\text{Subtract [3]:} & \text{-}\dfrac{(x-2)^2}{a^2} + \dfrac{y^2}{16} &=& \text{-}1 \end{array}$
$\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)} $
$\displaystyle \text{To have two intersections, the discriminant must be positive.}$
. . $\displaystyle Go\;for\;it!$