Locus problem (1 Viewer)

braintic · Jan 17, 2016

The locus of its centre ......... as what changes?

InteGrand · Jan 17, 2016

juantheron said:
$An ellipse with length of major axis $4$ and length of minor axis $2$$

$touches the coordinate axis , Then how can we prove that locus of$

$its center is circle $x^2+y^2 = 5$$

$\noindent (The question should say it touches the coordinate \textit{axes} (plural), i.e. it is tangent to both the $x$-axis and the $y$-axis.)$

$\noindent We will generalise it to an ellipse $\mathcal{E}$ with semi-major axis $a>0$ and semi-minor axis $b$, where $0<b\leq a$ (for this particular problem, $a=2$ and $b=1$).$

$\noindent Note that the locus must be symmetric about both the $x$-axis and the $y$-axis, due to the symmetry of an ellipse and of touching coordinate axes. This also means it will be symmetric about the lines $y=\pm x$.$

$\noindent We claim that the locus will not be the full circle $\mathcal{C}:x^2 + y^2 = a^2 + b^2$, but rather four arcs of this (one arc in each quadrant, by symmetry). The arc in the first quadrant will be the arc joining the point $(b,a)$ to $(a,b)$. Flip this arc about the $x$-axis, about the $y$-axis, and about the origin to obtain the other three arcs that form the locus. To see why we cannot have points with $|x|<b$ or $|y|<b$ on the circle $\mathcal{C}$ be part of the locus, note that if the centre of the ellipse is on such a point on the circle, it is too far away from one of the coordinate axes for the ellipse to touch it. For example, if the centre of the ellipse is in the first quadrant on $\mathcal{C}$ with $x<b$, then we must have $y>a$, since $x^2 + y^2 = a^2 + b^2$ on the ellipse. Then since the ellipse $\mathcal{E}$'s lowest point down from its centre can be $a$ units down (its semi-major axis), the ellipse cannot touch the $x$-axis.$

$\noindent For our proof, we will first show that the centre of the ellipse is indeed $\sqrt{a^2 + b^2}$ from the origin of the $x-y$ plane, and secondly, we will show that the arcs we mentioned above are precisely the attainable parts of the circle $\mathcal{C}$. For simplicity, we will consider everything occurring in the first quadrant, as we then obtain the other quadrants by symmetry.$

$\noindent Instead of considering things as $\mathcal{E}$ moves around in the $x-y$ plane, let's fix $\mathcal{E}$ in the Cartesian plane (call it the $X-Y$ plane, so $X$ on the horizontal axis and $Y$ on the vertical axis), and let's play play around with tangents of $\mathcal{E}$.$

$\noindent Let $\mathcal{E}$ be in the $X-Y$ plane in standard orientation, so its equation is $\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$. Let $P_1 \left(X_1,Y_1\right)$ be a point in the first quadrant on $\mathcal{E}$, with $0<X_1 < a$. Then the tangent to $\mathcal{E}$ at $P_1$ is $\mathcal{T}_1 : \frac{X_1 X}{a^2}+\frac{Y_1 Y}{b^2}=1$ (standard HSC formula).$

$\noindent We will find a point on $\mathcal{E}$ whose tangent $\mathcal{T}_2$ is perpendicular to $\mathcal{T}_1$. Then the tangents $\mathcal{T}_1$ and $\mathcal{T}_2$ are essentially the $x$- and $y$-axes of the $x-y$ plane, as the ellipse is tangent to these two perpendicular lines, and axes of a standard Cartesian plane are perpendicular. Thus the point of intersection of these two tangents can be considered the origin $O_{x-y}$ of the $x-y$ plane. So our goal will be to show that the distance from the centre of $\mathcal{E}$ (which is the origin $O_{X-Y}\equiv O$) of the $X-Y$ plane) is $\sqrt{a^2 + b^2}$. This will show that whenever the ellipse is tangent to both axes in the $x-y$ plane, its centre indeed lies on the circle $\mathcal{C}:x^2 + y^2 = a^2 + b^2$.$

$\noindent The tangent $\mathcal{T}_1 : \frac{X_1 X}{a^2}+\frac{Y_1 Y}{b^2}=1$ has slope $-\frac{b^2}{a^2}\cdot \frac{X_1}{Y_1}$. Let the perpendicular tangent $\mathcal{T}_2$ be tangent at the point $\left(X_2,Y_2\right)$ on $\mathcal{E}$. We seek to find $X_2$ and $Y_2$ in terms of $X_1$ and $Y_1$ (in fact there are two such points $\left(X_2,Y_2\right)$, reflections of each other about the origin, but we'll only need one of them for our purposes). Since $\mathcal{T}_2$ is perpendicular to $\mathcal{T}_1$, its slope needs to be $\frac{a^2}{b^2}\cdot \frac{Y_1}{X_1}$. Since the slope of the tangent at a point $\left(X_2,Y_2\right)$ on $\mathcal{E}$ is $-\frac{b^2}{a^2}\cdot \frac{X_2}{Y_2}$, we have$

$\begin{align*}-\frac{b^2}{a^2}\cdot \frac{X_2}{Y_2} &=\frac{a^2}{b^2}\cdot \frac{Y_1}{X_1} \\ \Rightarrow X_2 &= -\frac{a^4}{b^4}\cdot \frac{Y_1}{X_1}Y_2. \end{align*}$

$\noindent I'll start leaving out some algebraic details now, because things will get really tedious. I'll put them in a post below maybe. Solving the above with the equation of the ellipse $\mathcal{E}$, we find that we have $Y_2 = \pm \frac{b^4 X_1}{\sqrt{b^6 X_1 ^2 + a^6 Y_1 ^2}}$ and corresponding $X$ value $X_2 = \mp \frac{a^4 Y_1}{\sqrt{b^6 X_1^2 + a^6 Y_1 ^2}}$. We only need to worry about finding \textsl{a} point for $\mathcal{T}_2$ in order to construct the tangents that represent the $x$- and $y$-axes, so we take $\boxed{X_2 = \frac{a^4 Y_1}{\sqrt{b^6 X_1^2 + a^6 Y_1 ^2}}}$ and $\boxed{Y_2 = -\frac{b^4 X_1}{\sqrt{b^6 X_1 ^2 + a^6 Y_1 ^2}}}$.$

$\noindent Now, the equation of $\mathcal{T}_2$ is $\mathcal{T}_2:\frac{X_2 X}{a^2}+\frac{Y_2 Y}{b^2}=1$, where $X_2$ and $Y_2$ are defined above in terms of $X_1$ and $Y_1$. The origin of the $x-y$ plane $O_{x-y}$ is the point of intersection of these tangents. Solving these two tangent equations simultaneously gives us the point of intersection $\left(X_\text{int.},Y_{\text{int.}}\right)$ to be as follows:$

$\boxed{Y_{\text{int.}}=\frac{b^2 \left(X_1 - X_2\right)}{X_1 Y_2 - Y_1 X_2}}$

$\boxed{X_{\text{int.}}= \frac{a^2}{X_1 b^2} \left(b^2 - Y_1Y_{\text{int.}}\right)}.$

$\noindent Now let $d^2 \equiv X_{\text{int.}}^2 + Y_{\text{int.}}^2$ (this is the squared distance of the point of intersection of the tangents, i.e. $O_{x-y}$, to the centre of the ellipse $O$. Recall that our goal is to show that $d^2 = a^2 + b^2$. Now we have$

$$\begin{align*}d^2 \equiv X_{\text{int.}}^2 + Y_{\text{int.}}^2 &= \frac{a^4}{X_1 ^2 b^4} \left(b^2 - Y_1Y_{\text{int.}}\right)^2 +Y_{\text{int.}}^2 \quad (\text{using our boxed formula for } X_{\text{int.}}) \\ &= \frac{a^4}{X_1 ^2 b^4} \left(b^4 - 2b^2 Y_1Y_{\text{int.}} + Y_1^2 Y_{\text{int.}}^2 \right) +Y_{\text{int.}}^2 \\ &= \frac{a^4}{X_1 ^2}-\frac{2Y_1 Y_{\text{int.}}a^4}{b^2 X_1 ^2} + \frac{a^4 Y_1 ^2 Y_{\text{int.}}^2 }{b^4 X_1 ^2}+ Y_{\text{int.}}^2 \\ &= \frac{a^4}{X_1 ^2}-\frac{2Y_1a^4}{b^2 X_1 ^2}Y_{\text{int.}} + \left(\frac{a^4 Y_1^2 + b^4 X_1^2}{b^4 X_1^2} \right)Y_{\text{int.}}^2 \quad (\text{grouping }Y_{\text{int.}}^2 \text{ terms}) \\ &= \frac{a^4}{X_1 ^2}-\frac{2Y_1 a^4}{b^2 X_1 ^2}\cdot \frac{b^2 \left(X_1 - X_2\right)}{X_1 Y_2 - Y_1 X_2} + \left(\frac{a^4 Y_1^2 + b^4 X_1^2}{b^4 X_1^2} \right)\times \frac{b^4 \left(X_1 - X_2\right)^2}{\left(X_1 Y_2 - Y_1 X_2\right)^2}\quad (\text{using our boxed formula for }Y_\text{int.})\end{align*}$

$\begin{align*}= \frac{a^4}{X_1 ^2}-\frac{2Y_1 a^4}{ X_1 ^2}\cdot \frac{X_1 - X_2}{X_1 Y_2 - Y_1 X_2} + \frac{a^4 Y_1^2 + b^4 X_1^2}{X_1^2} \cdot \frac{\left(X_1 - X_2\right)^2}{\left(X_1 Y_2 - Y_1 X_2\right)^2}\end{align*}.$

$\noindent Now, it can be shown using our boxed formulas for $X_2$ and $Y_2$ that $X_1 Y_2 - Y_1 X_2 = -\frac{\left(b^4 X_1^2 + a^4 Y_1 ^2 \right)}{\sqrt{A}}$, where $A\equiv b^6 X_1 ^2 + a^6 Y_1^2$. So putting this in to our previous expression, we carry on to obtain$

$\begin{align*}d^2 &=\frac{a^4}{X_1 ^2}+\frac{2Y_1 a^4}{ X_1 ^2}\cdot \frac{X_1 - X_2}{b^4 X_1^2 + a^4 Y_1^2}\cdot \sqrt{A} + \left(\frac{X_1 - X_2}{X_1} \right)^2 \cdot \frac{A}{a^4 Y_1 ^2 + b^4 X_1^2} \\ &= \frac{a^4 \left(a^4 Y_1 ^2 + b^4 X_1^2 \right)+2Y_1 a^4 \cdot \left(X_1 - X_2\right) \sqrt{A}+ \left(X_1 - X_2\right)^2 \cdot A}{X_1 ^2 \left(a^4 Y_1 ^2 + b^4 X_1^2 \right)}. \end{align*}$

$\noindent This is equal to $a^2+b^2$ if and only if the numerator is equal to $\left(a^2 + b^2 \right)X_1^2 \left(a^4 Y_1 ^2 + b^4 X_1^2 \right)$, i.e. we now have to prove that$

$a^4 \left(a^4 Y_1 ^2 + b^4 X_1^2 \right)+2Y_1 a^4 \cdot \left(X_1 - X_2\right) \sqrt{A}+ \left(X_1 - X_2\right)^2 \cdot A=\left(a^2 + b^2 \right)X_1^2 \left(a^4 Y_1 ^2 + b^4 X_1^2 \right). \quad (**)$

$\noindent Note that$

$\begin{align*}\left(X_1 - X_2 \right)^2 &= X_1^2 -2X_1 X_2 + X_2^2 \\ &= X_1^2 - 2X_1 \cdot \frac{a^4 Y_1}{\sqrt{A}}+\frac{a^8Y_1^2}{A},\end{align*}$

$\noindent since from a boxed formula we have $X_2=\frac{a^4 Y_1}{\sqrt{A}}$, $A\equiv b^6 X_1^2 + a^6 Y_1 ^2$. Thus$

$\begin{align*}\left(X_1 - X_2 \right)^2 A &= \left(X_1^2 - 2X_1 \cdot \frac{a^4 Y_1}{\sqrt{A}}+\frac{a^8Y_1^2}{A} \right)A \\ &= X_1^2 A -2X_1 a^4 Y_1 \sqrt{A}+a^8 Y_1^2 \quad (\star). \end{align*}$

$Also,$

$\begin{align*}2Y_1 a^4 \left(X_1 - X_2 \right) \sqrt{A} &= 2Y_1 a^4 X_1 \sqrt{A} - 2Y_1 a^4 X_2 \sqrt{A} \\ &= 2Y_1 a^4 X_1 \sqrt{A} - 2Y_1 a^4 \cdot a^4 Y_1 \quad (\text{as }X_2 \sqrt{A} = a^4 Y_1) \\ &= 2Y_1 a^4 X_1 \sqrt{A} - 2Y_1 ^2 a^8.\quad (\star \star)\end{align*}$

$So putting the results of $(\star)$ and $(\star \star)$ into L.H.S. of $(**)$, we have$

$\begin{align*}LHS &= a^8 Y_1^2 + a^4 b^4 X_1^2 + 2Y_1 a^4 X_1 \sqrt{A} - 2Y_1^2 a^8 + X_1^2 A - 2X_1 a^4 Y_1 \sqrt{A} + a^8 Y_1^2 \\ &= a^4 b^4 X_1^2 + X_1^2 \cdot \left(b^6 X_1^2 + a^6 Y_1 \right) \quad (\text{as some terms cancel and replacing }A\text{ with what it is in terms of }X_1,Y_1) \\ &= X_1^2 \left(a^4 b^4 + b^6 X_1^2 + a^6 Y_1^2\right) \quad (\dagger).\end{align*}$

$The R.H.S. of RTP equation (**) is$

$\begin{align*}RHS &= \left(a^2 + b^2 \right)X_1 ^2 \left(a^4 Y_1^2 + b^4 X_1^2 \right) \\ &= X_1^2 \left(a^6 Y_1^2 + a^2b^4X_1^2 + b^2 a^4 Y_1^2 + b^6 X_1^2 \right) \quad (\ddagger) \end{align*}$

$\noindent Comparing this with $(\dagger)$, we see that (**) will be true iff $a^4 b^4 = a^2 b^4 X_1^2 + b^2 a^4 Y_1 ^2$. But this is indeed true, as it is just a rearrangement of the true statement $1=\frac{X_1^2}{a^2}+\frac{Y_1^2}{b^2}$, which is true because $\left(X_1,Y_1\right)$ lies on $\mathcal{E}$. So we have proved that the centre of the ellipse will always lie on the circle $\mathcal{C}$.$

$\noindent Finally, we prove our claim about the four arcs. We will just need to prove that the $x$-value of the locus takes on every value between $b$ and $a$ in the first quadrant (and no other value) \quad $(\diamond)$. From this it will follow that the locus in the first quadrant is the arc we mentioned, and the other arcs follow by symmetry. To prove claim $(\diamond)$, note that the $x$-value of the locus in the first quadrant as the ellipse varies is simply the perpendicular distance of the centre of the ellipse $\mathcal{E}$ (i.e. the origin in the $X-Y$ plane) from the tangent $\mathcal{T}_1$ (as this represents the positive $y$-axis say). This perpendicular distance $D$ is easily found using the tangent equation and the perpendicular distance formula. Since our point $P_1$ is in the first quadrant, we can parameterise it as $X_1 = a\cos \theta$, $Y_1 = b\sin \theta$, $\theta \in \left[ 0,\frac{\pi}{2}\right]$. Then the perpendicular distance is$

$\begin{align*}D&= \frac{|0+0 + (-1)|}{\sqrt{\frac{\cos^2 \theta}{a^2}+\frac{\sin^2 \theta}{b^2}}},\end{align*}$

$\noindent which simplifies to $D(\theta) = \frac{ab}{\sqrt{b^2 \cos^2 \theta + a^2 \sin^2 \theta}}$. Note that when $\theta = 0$, $D=a$ and when $\theta = \frac{\pi}{2}$, $D=b<a$. Also note that $D$ is monotone decreasing on $\left[ 0,\frac{\pi}{2}\right]$. This is because the function $f(\theta) := b^2 \cos^2 \theta + a^2 \sin^2 \theta$ is monotone \textit{increasing}, as it is equivalent to $b^2\left(1-\sin^2 \theta \right) + a^2 \sin^2 \theta = b^2 + \left(\underbrace{a^2 - b^2}_{>0}\right) \sin^2 \theta$, and this is clearly monotone increasing on this interval. Thus $D$ takes on values only between $b$ and $a$ in this interval (first quadrant of $\mathcal{E}$), and it takes on all these values by the intermediate value theorem, as it is a continuous function of $\theta$. Thus our claim is proved.$

$\noindent Some of our work in the first part of the proof may have relied on $P_1$ not having $X_1$ or $Y_1$ be 0, as then we may have had 0 denominators. But it is easy to see that in these cases, the distance in question is indeed $\sqrt{a^2 + b^2}$. Also, there is probably a nice geometric way to do this problem, but it may involve geometry theorems that aren't in the high school curriculum.$

Carrotsticks · Jan 19, 2016

It is (relatively) well-known that the director circle has equation x^2+y^2=a^2+b^2. I'll not prove this as it is a fairly standard textbook question.

As Integrand had suggested, we fix the ellipse and 'wrap' the perpendicular tangents (the XY axes) around the it. Let the intersection of tangents be T. The distance OT is always sqrt{a^2+b^2}.

There is a bijection between the fixed ellipse system (variable axes) and the variable ellipse (fixed axes) system up to scaling (we will keep the scaling to be 1:1). Hence in the variable ellipse system, the distance from the centre of the ellipse to the origin will be sqrt{a^2+b^2} for all positions of the ellipse satisfying the condition. In other words the locus of the centre is the circle x^2+y^2=a^2+b^2.

The restriction can be found similarly to above.

Locus problem (1 Viewer)

juantheron

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braintic

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InteGrand

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Carrotsticks

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