Parametrisation of Curves (1 Viewer)

RenegadeMx · Jul 28, 2016

InteGrand · Jul 28, 2016

RenegadeMx said:

$\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. $\boxed{y = \frac{8a^3}{x^2 + 4a^2}, -\infty<x<\infty }$ is the Cartesian equation ($x$ can be any real number since $x=2a\cot \phi$ and due to the range of values for $\phi$).$

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

RenegadeMx · Jul 28, 2016

InteGrand said:
$\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. $\boxed{y = \frac{8a^3}{x^2 + 4a^2}, -\infty<x<\infty }$ is the Cartesian equation (reason $x$ can be any real number is since $x=2a\cot \phi$ and due to the range of values for $\phi$).$

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

RenegadeMx · Sep 2, 2016

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

InteGrand · Sep 2, 2016

RenegadeMx said:
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 $y = x(c-x)$).$

RenegadeMx · Sep 2, 2016

InteGrand said:
$\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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Parametrisation of Curves (1 Viewer)

RenegadeMx

Kosovo is Serbian

InteGrand

Well-Known Member

RenegadeMx

Kosovo is Serbian

RenegadeMx

Kosovo is Serbian

InteGrand

Well-Known Member

RenegadeMx

Kosovo is Serbian

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