Page:Calculus Made Easy.pdf/110

The slope of the tangent must be the same as the ${\dfrac {dy}{dx}}$ of the curve; that is, $2x-5$.

The equation of the straight line is $y=ax+b$, and as it is satisfied for the values $x=2$, $y=-1$, then $-1=a\times 2+b$; also, its ${\dfrac {dy}{dx}}=a=2x-5$.

The $x$ and the $y$ of the point of contact must also satisfy both the equation of the tangent and the equation of the curve.

We have then

${\begin{cases}y&=x^{2}-5x+6,..................(i)\\y&=ax+b,.......................(ii)\\-1&=2a+b,.......................(iii)\\a&=2x-5,.......................(iv)\end{cases}}$

four equations in $a$, $b$, $x$, $y$.

Equations (i) and (ii) give $x^{2}-5x+6=ax+b$.

Replacing $a$ and $b$ by their value in this, we get

$x^{2}-5x+6=(2x-5)x-1-2(2x-5)$,

which simplifies to $x^{2}-4x+3=0$, the solutions of which are: $x=3$ and $x=1$. Replacing in (i), we get $y=0$ and $y=2$ respectively; the two points of contact are then $x=1$, $y=2$; and $x=3$, $y=0$.

Note.—In all exercises dealing with curves, students will find it extremely instructive to verify the deductions obtained by actually plotting the curves.

 

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