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Mathematics-Online problems:

Solution to the problem of the (previous) week

Problem:

The figure shows the parabola $ P: y = 4 - \dfrac{x^2}{3}$, that is tangent to the hyperbola $ H: y = \dfrac{a}{x}$ at $ B$ and intersects it at $ S$.


\includegraphics[width=.6\linewidth]{TdM_12_A1_bild}


Find the parameter $ a>0$ as well as the point of tangency $ B$ and the point of intersection $ S$ .


Answer:

$ a = $       
$ B = $ $ \big($ , $ \big)$  
$ S = $ $ \big($ , $ \big)$  

(The results should be correct to four decimal places.)

Solution:

Equating the slopes at the point of tangency, we obtain

$\displaystyle - \dfrac{a}{x^2} = - \dfrac{2}{3}x \Longleftrightarrow a = \dfrac{2}{3}x^3. $

Hence, we have

$\displaystyle \dfrac{\dfrac{2}{3}x^3}{{x}} = y = 4 - \dfrac{x^2}{3}\ . $

Simplifying, we get $ x^2 = 4$ and $ x = 2$, which yields

$\displaystyle a = \dfrac{16}{3} \approx 5.3333, \quad B = \left(2, \dfrac{8}{3}\right) \approx (2, 2.6667). $


Substitute $ y = a/x$ in the equation of the parabola to find the intersection point:

$\displaystyle \dfrac{16}{3x} = 4 - \dfrac{x^2}{3} \Longleftrightarrow p(x) = x^3 - 12x + 16 = 0 . $

Since $ x = 2$ is a zero of order 2 of $ p$ ($ x$-coordinate of the point of tangency), we can factor $ p$ and obtain the form

$\displaystyle p(x) = (x - 2)^2 (x - x_S) $

Substituting $ x=0$ in here, yields $ (-2)^2 (-x_S) = 16$. Thus, we get $ x_S = -4$ and

$\displaystyle S = \left(-4, -\dfrac{4}{3}\right) \approx (-4,-1.3333). $

[problem of the week]