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Worked solution

Eliminate exponential and trigonometric parameters

Find the Cartesian equation for each curve and give restrictions on xx where possible.

(i) x=etx=e^t, y=e3ty=e^{3t}.

(ii) x=ln⁡tx=\ln t, y=ty=\sqrt t, for t>0t>0.

(iii) x=5+4sec⁡θx=5+4\sec\theta, y=1+tan⁡θy=1+\tan\theta.

CJC · Discussion · Question 3

Answer

(i) y=x3y=x^3, x>0x>0. (ii) y=ex/2y=e^{x/2}, x∈Rx\in\mathbb R. (iii) (x−5)2/16−(y−1)2=1(x-5)^2/16-(y-1)^2=1, x≤1x\le1 or x≥9x\ge9.

(i) Use the exponential power rule

y=e3t=(et)3=x3y=e^{3t}=(e^t)^3=x^3, with x>0x>0.

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(ii) Solve for the positive parameter

t=ext=e^x, so y=ex=ex/2y=\sqrt{e^x}=e^{x/2}. Every real xx is possible; there is no further restriction on xx.

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(iii) Use the secant–tangent identity

sec⁡θ=(x−5)/4\sec\theta=(x-5)/4 and tan⁡θ=y−1\tan\theta=y-1. Since sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1, (x−5)2/16−(y−1)2=1(x-5)^2/16-(y-1)^2=1. The secant requires cos⁡θ≠0\cos\theta\ne0 and has values sec⁡θ≤−1\sec\theta\le-1 or sec⁡θ≥1\sec\theta\ge1. Hence x≤1x\le1 or x≥9x\ge9.

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Common mistakes

Keep the positive square root in (ii); the eliminated equation must respect the parameter range.

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