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Mathematics-Online problems:

Solution to the problem of the (previous) week


Problem:

The figure shows a quadrilateral bounded by two tangent lines to a parabola and their two normal lines. The parabola is described by the equation

$\displaystyle y=1-x^2 \,.
$


\includegraphics[width=0.5\linewidth]{TdM_08_A1_bild1.eps}


Determine the points $ P$ and $ Q$ as well as the area $ A$ of the shaded region.


Answer:

$ P = \Big($ , $ \Big)$
$ Q = \Big($ , $ \Big)$
$ A = $

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


Solution:

The straight line joining the center point and the point $ P = (x,1-x^2)$ has slope $ \left(1-x^2\right)/x$. This line is perpendicular to the tangent line which has slope $ -2x$. Thus, we have

$\displaystyle \left( -2x \right) \frac{1-x^2}{x}=-1\,.
$

Since $ x \neq 0$, we are allowed to divide by $ x$. Simplifying, we obtain

$\displaystyle 1-x^2=\frac{1}{2} \quad \Longleftrightarrow \quad
x^2= \frac{1}{2}\,.
$

Therefore,

$\displaystyle P= \left( \frac{\sqrt{2}}{2}\,,\,\frac{1}{2} \right) \approx (0.7071\
,\,0.5) \,.
$

From the point-slope form of the tangent line,

$\displaystyle \frac{y-1/2}{x-\sqrt{2}/2}= -2 \, \frac{\sqrt{2}}{2} \,,
$

we obtain with $ x = 0$

$\displaystyle y=\frac{1}{2} +2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} =
\frac{3}{2} = 1.5
$

for the $ y$-coordinate of point $ Q$.

The area $ A$ is half the product of the lengths of the diagonals:

$\displaystyle A = \frac{1}{2} \cdot 2 \cdot \frac{\sqrt{2}}{2} \cdot
\frac{3}{2} = \frac{3}{4} \, \sqrt{2} \approx 1.0607 \,.
$


[problem of the week]