Page:Calculus Made Easy.pdf/109

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

${\begin{cases}y=2x^{2}+2,\\y=-{\tfrac {1}{2}}x+2\quad {\text{or}}\quad 2x^{2}+2=-{\tfrac {1}{2}}x+2;\end{cases}}$

that is,

$x(2x+{\tfrac {1}{2}})=0$.

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

${\dfrac {dy}{dx}}=4x$.

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

$x=-{\dfrac {1}{4}},\quad {\dfrac {dy}{dx}}=-1;$

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

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

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