Question

Investigation \(\quad\) Find all \(a\) and \(b\) such that the two curves given by \(y_{1}=a x(b-x)\) and \(y_{2}=\frac{x}{x+2}\) intersect at only one point and have a common tangent line and equal curvature at that point. Sketch a graph for each set of values of \(a\) and \(b\).

Short answer

The values of \(a\) and \(b\) can be found by solving the three equations obtained from the condition of intersection, equal tangent lines and curvatures. Then the graphs can be sketched using these values.

Step-by-step solution

Setup Intersection Equality

From the description, both curves intersect at a single point. This implies that at that point, the y-values from both equations are equal i.e. \(y_{1}=y_{2}\) or \(a x(b-x)=\frac{x}{x+2}\). This gives the first equation.

Setup Tangent Line Equality

The second condition says that both curves share a common tangent line at the point of intersection. The slope of the tangent line is the derivative of the function. For \(y_{1}\), \(\frac{dy_{1}}{dx} = ab - 2ax\). For \(y_{2}\), \(\frac{dy_{2}}{dx} = \frac{2}{(2+x)^2}\). So, \(\frac{dy_{1}}{dx}= \frac{dy_{2}}{dx}\) or \(ab - 2ax = \frac{2}{(2+x)^2}\). This gives the second equation.

Setup Curvature Equality

The third condition says that both curves have the same curvature at the point of intersection. Curvature is given by the second derivative of the function. For \(y_{1}\), \(\frac{d^2y_{1}}{dx^2} = -2a\). For \(y_{2}\), \(\frac{d^2y_{2}}{dx^2} = \frac{-4}{(2+x)^3}\). So, \(\frac{d^2y_{1}}{dx^2}= \frac{d^2y_{2}}{dx^2}\) or \(-2a = \frac{-4}{(2+x)^3}\). This gives the third equation.

Solve Equations

Solving these three equations simultaneously would yield the values of \(a\) and \(b\).

Sketching the Graphs

The final step involves sketching the graph for each pair of \(a\) and \(b\) solutions. Use the resulting coordinates from the solution in the equations of the lines to plot the points and draw the curves.