Pure Mathematics 2

Mathematics9709

Pure Mathematics 2

Mathematics9709

Parametric Tangents and Normals

A parameter value is not a coordinate. Use it three times: to find the point, to find the tangent gradient, and then to build the required tangent or normal.

Point and gradient

Substitute the given parameter into $x(t)$ and $y(t)$ to find the point.
Find $dx/dt$ and $dy/dt$ , then evaluate their ratio.
For a normal, use the negative reciprocal $m_n=-1/m_t$ .
Use $y-y_1=m(x-x_1)$ and convert to the requested form.

Exact current-paper example

Keep exact trig values until the line equation is complete.

Worked example

A current-paper normal

x=4\cos^2t

y=\sqrt3\sin2t

\frac{dx}{dt}=-4\sin2t

\frac{dy}{dt}=2\sqrt3\cos2t

At $t=\pi/6$ , the point is $(3,3/2)$ and $dy/dx=-1/2$ . The normal gradient is 2.

y-\frac32=2(x-3)

4x-2y-9=0

(9709/22/M/J/24 Q4)

Common mistake

A normal gradient is the negative reciprocal, not only the negative. For example, the negative reciprocal of

-1/2

is 2.

Build the required line

Your turnIndependent check [6 marks]

For $x=t^2+1$ and $y=t^3-3t$ , find the normal at $t=2$ in the form $ax+by+c=0$ .

Show worked answer

\frac{dx}{dt}=2t,\quad\frac{dy}{dt}=3t^2-3

At $t=2$ , the point is $(5,2)$ , the tangent gradient is $9/4$ and the normal gradient is $-4/9$ .

Answer

4x+9y-38=0

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