Parametric curve to Cartesian form and tangent

Parametric equations

A-level Maths (9MA0)

ORIGINAL

Eliminate the parameter, then use parametric differentiation to find a tangent on the same curve.

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Question

A curve has parametric equations

x=t+\dfrac1t

and

y=t-\dfrac1t

, where

t>0

. Find a Cartesian equation of the curve and the equation of the tangent when

t=2

Every step worked, with the reasoning.

1
$x+y=2t$ and $x-y=\dfrac{2}{t}$
Add and subtract the parametric equations to isolate $t$ and $1/t$ .
2
$(x+y)(x-y)=4$ , so $x^2-y^2=4$ , with $x\ge2$
Multiply to eliminate $t$ . Since $t>0$ , $x=t+1/t\ge2$ , so only the right-hand branch is traced.
3
$\dfrac{dx}{dt}=1-\dfrac1{t^2}$ and $\dfrac{dy}{dt}=1+\dfrac1{t^2}$
Differentiate both parametric equations with respect to $t$ .
4
$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{1+1/t^2}{1-1/t^2}$
Divide the two parameter derivatives.
5
At $t=2$ , $(x,y)=(\tfrac52,\tfrac32)$ and $\dfrac{dy}{dx}=\dfrac{5/4}{3/4}=\dfrac53$ .
Find the point and gradient at the specified parameter value.
6
$y-\dfrac32=\dfrac53\left(x-\dfrac52\right)$ , so $5x-3y-8=0$
Use the point-gradient equation and simplify.

Answer: $x^2-y^2=4$ with $x\ge2$ ; tangent $5x-3y-8=0$ .