that is, it must be a solution of the system of simultaneous

equations formed by coupling together the equations of the curves. Here the curves meet one another at points given by the solution of

that is,

This equation has for its solutions ${\displaystyle x=0}$ and ${\displaystyle x=-{\tfrac {1}{4}}}$. The slope of the curve ${\displaystyle y=2x^{2}+2}$ at any point is

For the point where ${\displaystyle x=0}$, this slope is zero; the curve is horizontal. For the point where

hence the curve at that point slopes downwards to the right at such an angle ${\displaystyle \theta }$ with the horizontal that ${\displaystyle \tan \theta =1}$; that is, at ${\displaystyle 45^{\circ }}$ to the horizontal.

The slope of the straight line is ${\displaystyle -{\tfrac {1}{2}}}$; that is, it slopes downwards to the right and makes with the horizontal an angle ${\displaystyle \phi }$ such that ${\displaystyle \tan \phi ={\tfrac {1}{2}}}$; that is, an angle of ${\displaystyle }$. It follows that at the first point the curve cuts the straight line at an angle of ${\displaystyle 26^{\circ }34^{\prime }}$, while at the second it cuts it at an angle of ${\displaystyle 45^{\circ }-26^{\circ }34^{\prime }=18^{\circ }26^{\prime }}$.

(5) A straight line is to be drawn, through a point whose coordinates are ${\displaystyle x=2}$, ${\displaystyle y=-1}$, as tangent to the curve ${\displaystyle y=x^{2}-5x+6}$. Find the coordinates of the point of contact.