Q.The abscissa of the point on the curve $f(x)=\sqrt{8-2x}$ at which the slope of the tangent is $-0.25$ ?

Tangents and Normals to a Curve

At a point $P(x_1,y_1)$ on $y=f(x)$ with slope $m=\dfrac{dy}{dx}$, the tangent line is

$y-y_1=m(x-x_1)$ and the normal has slope $-\dfrac1m$. Four classical lengths are

measured from $P$ and the foot $A$ of the ordinate on the $x$-axis:

• Length of tangent $=\Big|\dfrac{y_1\sqrt{1+m^2}}{m}\Big|$,
• Length of normal $=\big|y_1\sqrt{1+m^2}\big|$,
• Subtangent $ST=\Big|\dfrac{y_1}{m}\Big|$,
• Subnormal $SN=|m\,y_1|$.

From these, $SN\cdot ST=y_1^2$ and $\dfrac{SN}{ST}=m^2$. For a curve given

parametrically ($x=x(\theta),y=y(\theta)$) the slope is $\dfrac{dy/d\theta}{dx/d\theta}$

and the same length formulas apply. These quantities let one compare tangent, normal, …