A quarter ellipse becomes a self-inverse circle

Sketch the curve y2+ax2=5 y^2+ax^2=5 for x≥0 x\ge0, y≥0 y\ge0, where aa is positive. mark

f:x↦5−ax2 f:x\mapsto\sqrt{5-ax^2}, x∈Rx\in\mathbb R, 0≤x≤5/a0\le x\le\sqrt{5/a}; g:x↦1+e−x g:x\mapsto1+e^{-x}, x∈Rx\in\mathbb R, x≥0 x\ge0.

Show that f−1 f^{-1} exists and define f−1 f^{-1} in a similar form. marks

Given f2(x)=xf^2(x)=x for all x∈Rx\in\mathbb R, 0≤x≤5/a0\le x\le\sqrt{5/a},

(i) show a=1 a=1 without evaluating f2(x) f^2(x), marks

(ii) using (i), show fgfg exists and find its corresponding range. marks

[2011/ACJC/Prelim/I/9]

Total marks: 10

Solution

Sketch

Sketching conics

Sketch y2+ax2=5y^2+ax^2=5 for x≥0x\ge0, y≥0y\ge0, where a>0a>0.

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Hint

  1. Divide by 5: y25+x25/a=1\frac{y^2}5+\frac{x^2}{5/a}=1 is an ellipse.

  2. Find the intercepts and keep the first quadrant.

First quadrant ellipse
First quadrant ellipseScroll across to see the full diagram.

Answer · Sketch

Quarter-ellipse endpoints (0,5),(5/a,0)(0,\sqrt5),(\sqrt{5/a},0).

a>0a>0, x,y≥0.

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Open interactive exploration

The quarter-ellipse changes with a

The vertical intercept stays at √5. The horizontal intercept √(5/a) shrinks as a grows.

Teal: a = 0.5. No comparison overlay.

Start: a = 0.5

a=0.5a=0.5: intercepts (0,5)(0,\sqrt5) and (10,0)≈(3.16,0)(\sqrt{10},0)\approx(3.16,0).

  • Quarter-ellipse
  • Intercepts
Ideas to try
Which value of aa makes the curve look the same after swapping xx and yy?