2. ## Re: Parametrisation of Curves

$\noindent It is clear that the locus will be symmetric about the y-axis, so if a point (x,y) is on the locus, so is (-x,y). So let's just focus on when x\geq 0 (i.e. when B lies on or to the right of the y-axis).$

$\noindent Note that the polar equation of the circle in equation is r=2a\sin \theta (standard polar equation for such a circle). Let the angle between the tangent L and OA be \phi, 0<\phi \leq \frac{\pi}{2}. By this I am referring to \angle OAX, where X is the point on the top of the circle, i.e. (0,2a). The y-value of P, y_P, is just the y-value of B. Since OB is of length 2a\sin \phi (from the polar equation of the circle), and the angle made by OB with the positive x-axis is \phi (using alternate angles in parallel lines), we have that the y-value of B is r\sin\phi = 2a\sin \phi\cdot \sin \phi = 2a\sin^{2}\phi. So y_P = 2a\sin^{2}\phi.$

$\noindent For the x-value of P, x_P, it is just the length XA. From right-angle trigonometry in triangle OXA, noting OX = 2a, we have XA = 2a\cot \phi. So x_P= 2a\cot \phi.$

$\noindent Thus using symmetry, to get the full locus, when B is on the left of the y-axis, we can treat the angle \phi as negative (e.g. -30^\circ), and this will negate x_P whilst leaving y_P unchanged, as we want. So the parametric equations of the locus in this setup are \boxed{x=2a\cot\phi, y = 2a\sin^{2}\phi,\text{ for }\phi \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right]\backslash \left\{0\right\}}. Note that when \phi = 90^\circ = \frac{\pi}{2}, we have a degenerate' case with A and B both at (0,2a), and P also at (0,2a). The parametric equations still hold when \phi = 90^\circ though.$

$\noindent To get an implicit equation from x=2a\cot\phi, y = 2a\sin^{2}\phi for \phi \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right]\backslash \left\{0\right\} we use trig. identities. Note y can never be 0 (both seen geometrically and from our restriction on \phi). Therefore, we can write \csc^{2}\phi = \frac{2a}{y} by rearranging y = 2a\sin^{2}\phi. Since \cot^{2}\phi+1 = \csc^{2}\phi, and \cot\phi = \frac{x}{2a}, we have \frac{x^2}{4a^2}+1 = \frac{2a}{y} as an implicit Cartesian equation. But we can easily solve this for y, i.e. x^2 y + 4a^2 y = 8a^3, i.e. \boxed{y = \frac{8a^3}{x^2 + 4a^2}, -\infty

For anyone curious, there's further info on the Witch of Agnesi (with an animation too) here: http://mathworld.wolfram.com/WitchofAgnesi.html .

3. ## Re: Parametrisation of Curves

Originally Posted by InteGrand
$\noindent It is clear that the locus will be symmetric about the y-axis, so if a point (x,y) is on the locus, so is (-x,y). So let's just focus on when x\geq 0 (i.e. when B lies on or to the right of the y-axis).$

$\noindent Note that the polar equation of the circle in equation is r=2a\sin \theta (standard polar equation for such a circle). Let the angle between the tangent L and OA be \phi, 0<\phi \leq \frac{\pi}{2}. The y-value of P, y_P, is just the y-value of B. Since OB is of length 2a\sin \phi (from the polar equation of the circle), and the angle made by OB with the positive x-axis is \phi (using alternate angles in parallel lines), we have that the y-value of B is r\sin\phi = 2a\sin \phi\cdot \sin \phi = 2a\sin^{2}\phi. So y_P = 2a\sin^{2}\phi.$

$\noindent For the x-value of P, x_P, it is just the length XA, where X is the point on the top of the circle, i.e. (0,2a). From right-angle trigonometry in triangle OXA, noting OX = 2a, we have XA = 2a\cot \phi. So x_P= 2a\cot \phi.$

$\noindent Thus using symmetry, to get the full locus, when B is on the left of the y-axis, we can treat the angle \phi as negative (e.g. -30^\circ), and this will negate x_P whilst leaving y_P unchanged, as we want. So the parametric equations of the locus in this setup are \boxed{x=2a\cot\phi, y = 2a\sin^{2}\phi,\text{ for }\phi \in \left(-\frac{\pi}{2},\right]\backslash \left\{0\right\}}. Note that when \phi = 90^\circ = \frac{\pi}{2}, we have a degenerate' case with A and B both at (0,2a), and P also at (0,2a). The parametric equations still hold when \phi = 90^\circ though.$

$\noindent To get an implicit equation from x=2a\cot\phi, y = 2a\sin^{2}\phi for \phi \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right]\backslash \left\{0\right\} we use trig. identities. Note y can never be 0 (both seen geometrically and from our restriction on \phi). Therefore, we can write \csc^{2}\phi = \frac{2a}{y} by rearranging y = 2a\sin^{2}\phi. Since \cot^{2}\phi+1 = \csc^{2}\phi, and \cot\phi = \frac{x}{2a}, we have \frac{x^2}{4a^2}+1 = \frac{2a}{y} as an implicit Cartesian equation. But we can easily solve this for y, i.e. x^2 y + 4a^2 y = 8a^3, i.e. \boxed{y = \frac{8a^3}{x^2 + 4a^2}, -\infty

For anyone curious, there's further info on the Witch of Agnesi (with an animation too) here: http://mathworld.wolfram.com/WitchofAgnesi.html .
very impressive, ty

4. ## Re: Parametrisation of Curves

found an old class test, how would one go about doing this, i can sorta see that the eqn will be something y=-x^2 + extras but otherwise am stuck

5. ## Re: Parametrisation of Curves

found an old class test, how would one go about doing this, i can sorta see that the eqn will be something y=-x^2 + extras but otherwise am stuck

$\noindent Let B be the origin and C = (c,0) on the x-axis, where c is a fixed positive number. By a suitable choice of units and since whether the line L is above or below the x-axis clearly won't affect the \emph{shape} of the locus, we can just take L to be the line y = 1. Now, let A be the point (t,1) on this line, where t\in \mathbb{R} will be allowed to vary.$

$\noindent Then the altitude from A to BC is just the vertical line through A, i.e. the line x = t. Note for t\neq c, the slope of AC is \frac{1-0}{t-c} = \frac{1}{t-c}. So the slope of the altitude from B to AC is the negative reciprocal of this, i.e. c-t (this is also true for t=c, since then, A is directly above C, so the line BC \emph{is} the altitude to AC, and has slope 0). So the equation of the altitude from B to AC is y = (c-t)x (since B is the origin).$

$\noindent So the intersection of these two altitudes (which is precisely the orthocentre) is at the point where x = t and y = (c-t)t. So there are the parametric equations of the orthocentre, which clearly gives us a parabola (which Cartesian equation y = x(c-x)).$

6. ## Re: Parametrisation of Curves

Originally Posted by InteGrand
$\noindent Let B be the origin and C = (c,0) on the x-axis, where c is a fixed positive number. By a suitable choice of units and since whether the line L is above or below the x-axis clearly won't affect the \emph{shape} of the locus, we can just take L to be the line y = 1. Now, let A be the point (t,1) on this line, where t\in \mathbb{R} will be allowed to vary.$

$\noindent Then the altitude from A to BC is just the vertical line through A, i.e. the line x = t. Note for t\neq c, the slope of AC is \frac{1-0}{t-c} = \frac{1}{t-c}. So the slope of the altitude from B to AC is the negative reciprocal of this, i.e. c-t (this is also true for t=c, since then, A is directly above C, so the line BC \emph{is} the altitude to AC, and has slope 0). So the equation of the altitude from B to AC is y = (c-t)x (since B is the origin).$

$\noindent So the intersection of these two altitudes (which is precisely the orthocentre) is at the point where x = t and y = (c-t)t. So there are the parametric equations of the orthocentre, which clearly gives us a parabola (which Cartesian equation y = x(c-x)).$
ahh forgot to realise altitude from A to BC will always have the same x eqn, anyways thx again

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