Porisma

Worked solution

Retain the positive branch when eliminating a radical parameter

For x=1/t+1x=1/\sqrt{t+1} and y=t/(t+1)y=t/(t+1), find a Cartesian equation and sketch the curve.

CJC · Lecture · Example 4

Isolate t using y

y(t+1)=ty(t+1)=t gives t(1−y)=yt(1-y)=y, hence t=y/(1−y)t=y/(1-y). Therefore x=1/y/(1−y)+1=1/1/(1−y)=1−yx=1/\sqrt{y/(1-y)+1}=1/\sqrt{1/(1-y)}=\sqrt{1-y}.

How did this part go?

Restore the restrictions from the original equations

The real reciprocal square root requires t>−1t>-1, and x is strictly positive. Thus y=1−x2y=1-x^2, x>0x>0. This is only the right half of the downward parabola: it approaches the excluded point (0,1)(0,1) as t→∞t\to\infty and crosses the x-axis at (1,0)(1,0) when t=0t=0. As t→−1+t\to-1^+, x tends to infinity and y to minus infinity.

How did this part go?

Compare your sketch

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The right-hand branch y=1−x² for x>0, open circle (0,1), included intercept (1,0), unbounded downward tail; increasing t moves toward the open endpoint.
The right-hand branch y=1−x² for x>0, open circle (0,1), included intercept (1,0), unbounded downward tail; increasing t moves toward the open endpoint.
The right-hand branch y=1−x² for x>0, open circle (0,1), included intercept (1,0), unbounded downward tail; increasing t moves toward the open endpoint.

Common mistakes

The Cartesian polynomial alone would add the forbidden left branch and the endpoint x=0.

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