The witch of Maria Agnesi

[ali PMPAINT]

Homework Statement

Give the Parametric Equation of it.

Relevant Equations

x = AQ, y = 2 - ABsin(t), AB*OA = AQ^2

So, I understood the solution and the method, but I didn't get one part: How did it conclude that AB*OA = (AQ)^2 ??
I tried to derive it using similarities(with ABP and AQO), but no luck. Then, I tried to connect B to the origin of the circle(call it R) and forming BOR, and obtaining OB, and although the Parametric Equation of it can be derived, it would be much more complex compared to the book solution and it doesn't conclude AB*OA = (AQ)^2 . I don't know what to do next, any suggestions?

 

[tnich]

Try drawing the line segment BQ. What is the angle OBQ?

 

[Michael Price]

The trick is to see that OA * OB = 4. One way is use similar triangles, QAO and QOB.

 

[aheight]

This is the diagram I used. If anyone is interested, I posted Mathematica code below for an interactive Mainpulate. In the diagram, can we not say:
(corrected following as per comment below)
$$
\triangle QAO \sim \triangle BQO \sim \triangle BAQ
$$
via AAS rule. Then we have:

$$
\frac{QA}{BA}=\frac{AO}{AQ}
$$
or $$(QA)^2=AO\cdot BA?$$

theCircle =
  ContourPlot[x^2 + (y - 1)^2 == 1, {x, -3, 3}, {y, -3, 3},
   PlotRange -> 5, Axes -> True];
theLine = Graphics[Line[{{-3, 2}, {3, 2}}]];
theX[t_] := 2 Cot[t];
theY[t_] := 2 Sin[t]^2
myT = Pi/4;
witchPt = ParametricPlot[{theX[t], theY[t]}, {t, Pi/14, Pi/2}];
Manipulate[
point1 = Graphics[{Red, PointSize[0.01], Point[theA]}];
point2 =
  Graphics[{Blue, PointSize[0.01], Point[{theX[myT], theY[myT]}]}];
bLen[t_] := {theX[t] - 2 Sin[t] Cot[t]^2 Cos[t], theY[t]};
myLine2 = Graphics[Line[{{theX[myT], theY[myT]}, bLen[myT]}]];
myLine = Graphics[Line[{{0, 0}, {theX[myT], 2}}]];
myLine3 = Graphics[Line[{{theX[myT], theY[myT]}, {theX[myT], 2}}]];
greenPoint =
  Graphics[{Darker@Green, PointSize[0.01], Point[bLen[myT]]}];
myLine4 = Graphics[{Darker@Green, Line[{{0, 2}, bLen[myT]}]}];
theLabels = Graphics[{Style[Text["Q", {-0.2, 2.2}], 14],
    Style[Text["A", {theX[myT], 2.2}], 14],
    Style[Text["P", {theX[myT] + 0.2, theY[myT] + 0.1}], 14],
    Style[Text["B", bLen[myT] - {-0.1, 0.15}], 14],
    Style[Text["O", {-0.15, -0.15}], 14]}];
Show[{theCircle, theLine, witchPt, myLine, point1, point2, myLine2,
   myLine3, myLine4, theLabels, greenPoint}], {myT, Pi/20, Pi/2},
TrackedSymbols :> True]

 

[Michael Price]

Should that be BAQ, not BQA?

 

[aheight]

Should that be BAQ, not BQA?

Ok thanks, I'll fix it.

 

[ali PMPAINT]

This is the diagram I used. If anyone is interested, I posted Mathematica code below for an interactive Mainpulate. In the diagram, can we not say:
(corrected following as per comment below)
$$
\triangle QAO \sim \triangle BQO \sim \triangle BAQ
$$
via AAS rule. Then we have:

$$
\frac{QA}{BA}=\frac{AO}{AQ}
$$
or $$(QA)^2=AO\cdot BA?$$
View attachment 248428

theCircle =
  ContourPlot[x^2 + (y - 1)^2 == 1, {x, -3, 3}, {y, -3, 3},
   PlotRange -> 5, Axes -> True];
theLine = Graphics[Line[{{-3, 2}, {3, 2}}]];
theX[t_] := 2 Cot[t];
theY[t_] := 2 Sin[t]^2
myT = Pi/4;
witchPt = ParametricPlot[{theX[t], theY[t]}, {t, Pi/14, Pi/2}];
Manipulate[
point1 = Graphics[{Red, PointSize[0.01], Point[theA]}];
point2 =
  Graphics[{Blue, PointSize[0.01], Point[{theX[myT], theY[myT]}]}];
bLen[t_] := {theX[t] - 2 Sin[t] Cot[t]^2 Cos[t], theY[t]};
myLine2 = Graphics[Line[{{theX[myT], theY[myT]}, bLen[myT]}]];
myLine = Graphics[Line[{{0, 0}, {theX[myT], 2}}]];
myLine3 = Graphics[Line[{{theX[myT], theY[myT]}, {theX[myT], 2}}]];
greenPoint =
  Graphics[{Darker@Green, PointSize[0.01], Point[bLen[myT]]}];
myLine4 = Graphics[{Darker@Green, Line[{{0, 2}, bLen[myT]}]}];
theLabels = Graphics[{Style[Text["Q", {-0.2, 2.2}], 14],
    Style[Text["A", {theX[myT], 2.2}], 14],
    Style[Text["P", {theX[myT] + 0.2, theY[myT] + 0.1}], 14],
    Style[Text["B", bLen[myT] - {-0.1, 0.15}], 14],
    Style[Text["O", {-0.15, -0.15}], 14]}];
Show[{theCircle, theLine, witchPt, myLine, point1, point2, myLine2,
   myLine3, myLine4, theLabels, greenPoint}], {myT, Pi/20, Pi/2},
TrackedSymbols :> True]

Ah, now it's clear to me. Thanks for the help.

 

[ali PMPAINT]

Should that be BAQ, not BQA?

What is the difference?

 

[Michael Price]

What is the difference?

The angle vertex is normally located at the middle letter