2016ers Chit-Chat Thread (4 Viewers)

DatAtarLyfe · Mar 25, 2016

That somewhat makes sense.

InteGrand · Mar 25, 2016

For that ellipse question, the y-value of the circle's centre is clearly 0 by symmetry (specifically it must be equidistant from the two given points on the ellipse, which means it lies on the perpendicular bisector to the line segment joining those two points, which is the x-axis).

A way to find the x-value (in fact, the entire coordinates) of the centre is as leehuan was saying, the centre will be the point of intersection of the normals at those points on the ellipse, so we could find equations of both normals and solve simultaneously. However, since we already argued above that y = 0 at the circle centre, we only need to actually find one normal equation (and don't need to do simultaneous solving). Then just find the x-intercept of it (i.e. sub. y = 0 and solve for x). This gives us the x-value of the coordinates of the circle's centre, with the y-value being 0.

leehuan · Mar 25, 2016

Even when tired I can do maths to a valid albeit poor extent. :')

DatAtarLyfe · Mar 25, 2016

Thanks guys, just learned something new

Nailgun · Mar 25, 2016

InteGrand said:
For that ellipse question, the y-value of the circle's centre is clearly 0 by symmetry (specifically it must be equidistant from the two given points on the ellipse, which means it lies on the perpendicular bisector to the line segment joining those two points, which is the x-axis).

A way to find the x-value (in fact, the entire coordinates) of the centre is as leehuan was saying, the centre will be the point of intersection of the normals at those points on the ellipse, so we could find equations of both normals and solve simultaneously. However, since we already argued above that y = 0 at the circle centre, we only need to actually find one normal equation (and don't need to do simultaneous solving). Then just find the x-intercept of it (i.e. sub. y = 0 and solve for x). This gives us the x-value of the coordinates of the circle's centre, with the y-value being 0.

Hey this is kind of what I tried doing - but it didn't work lol

leehuan · Mar 25, 2016

Nailgun said:
Hey this is kind of what I tried doing - but it didn't work lol

$\sum _{ maths=method }^{ arithmetic }{ Nailgun } =Confusexplosion$

Nailgun · Mar 25, 2016

nailgun said:
hey this is kind of what i tried doing - but it didn't work lol

WAIT WAIT

I GOT THE ANSWER
MY METHOD WAS LEGIT
I JUST MADE A MISTAKE

fullsrs

InteGrand · Mar 25, 2016

Nailgun said:
wait wait wait
my method was legit
i just made a mistake
full srs

q is really easy

What was your method? Finding both normals and solving them simultaneously? (By the way, I realised now that getting the equation of one of the normals was an earlier part of the question.)

leehuan · Mar 25, 2016

HAHAHAHA.

I knew I was right to put arithmetic on the top of my summation joke

leehuan · Mar 25, 2016

InteGrand said:
What was your method? Finding both tangents and solving them simultaneously? (By the way, I realised now that getting the equation of one of the tangents was an earlier part of the question.)

I thought the question only wanted the normal

DatAtarLyfe · Mar 25, 2016

Now i feel stupid

InteGrand · Mar 25, 2016

leehuan said:
I thought the question only wanted the normal

Oh sorry, I meant normal in all instances above, not tangent.

DatAtarLyfe · Mar 25, 2016

Couldnt you have simultaneous the normals at both points?
Edit: k cool

leehuan · Mar 25, 2016

DatAtarLyfe said:
Couldnt you have simultaneous the normals at both points?

You can but the deduction InteGrand presented allows you to avoid finding the equation of the second normal.

Nailgun · Mar 25, 2016

InteGrand said:
What was your method? Finding both tangents and solving them simultaneously? (By the way, I realised now that getting the equation of one of the tangents was an earlier part of the question.)

idk how you found y=0 by symmetry

if P and Q are on the circle, then the distance between P and the center (h,k) will be radius

so r^2 = (5-h)^2+(7.5-k)^2

equation of circle is (x-h)^2-(y-k)^2 = (5-h)^2+(7.5-k)^2
sub x = 5 y = -7.5

k = 0

therefore y coordinate is 0

then if circle is tangent to ellipse
the equation of tangent at x=5 will be the same for both ellipse and circle
if the tangents are the same the normals are the same
the normal of a tangent of a circle always passes through centre ( i think this is a circle geo thing)
so sub y=0 into normal they give you

DatAtarLyfe · Mar 25, 2016

How nailgun did it originally was my original thought but i figured there had to be a reason they made me find the normal in part ii

leehuan said:
You can but the deduction InteGrand presented allows you to avoid finding the equation of the second normal.

Q is simply a reflection of P about the x-axis. So i couldve just modified the normal in part ii

InteGrand · Mar 25, 2016

Nailgun said:
idk how you found y=0 by symmetry

if P and Q are on the circle, then the distance between P and the center (h,k) will be radius

so r^2 = (5-h)+(7.5-k)^2

equation of circle is (x-h)^2-(y-k)^2 = (5-h)+(7.5-k)^2
sub x = 5 y = -7.5

k = 0

therefore y coordinate is 0

then if circle is tangent to ellipse
the equation of tangent at x=5 will be the same
so turn the normal eq they give you into a tangent
sub y=0
profit

$\noindent There are a few ways to argue that $y=0$ by symmetry. The first is to draw in the circle with centre $(h,k)$ and the ellipse $E$, with the two points $P$ and $Q$ labelled. Then note that if we applied a reflection transformation $\mathcal{R}$ to the plane so that everything is reflected about the $x$-axis, we have $P$ go to where $Q$ was, $Q$ go to where $P$ was, and the centre of the circle is moved to $(h, -k)$. But the graph of the ellipse $E$ is invariant under $\mathcal{R}$, because it is symmetric about the $x$-axis.$

$\noindent This implies (as mentioned before), that $P$ and $Q$ stay in the same `places', just with their names switched around. The circle is still tangent to $P$ and $Q$ after the reflection (because applying a reflection to the whole plane doesn't suddenly stop things from being tangent), which means that it is still tangent at $\left(5,7\tfrac{1}{2}\right)$ and $\left(5,-7\tfrac{1}{2}\right)$ in the reflected plane. This means the circle is still the same circle as before (because it is still tangent to the same ellipse (as the ellipse was unchanged) at the same points), so its centre is the same, so $(h,-k)$ must be the same as $(h,k)$, which means $k=-k\Rightarrow k = 0$.$

$\noindent Another way to see that $k$ must be 0 is that as the circle's centre is equidistant from $P$ and $Q$, it must lie on the perpendicular bisector of $P$ and $Q$, which means it lies on the $x$-axis. Hence $k=0$.$

Nailgun · Mar 25, 2016

InteGrand said:
$\noindent There are a few ways to argue that $y=0$ by symmetry. The first is to draw in the circle with centre $(h,k)$ and the ellipse $E$, with the two points $P$ and $Q$ labelled. Then note that if we applied a reflection transformation $\mathcal{R}$ to the plane so that everything is reflected about the $x$-axis, we have $P$ go to where $Q$ was, $Q$ go to where $P$ was, and the centre of the circle is moved to $(h, -k)$. But the graph of the ellipse $E$ is invariant under $\mathcal{R}$, because it is symmetric about the $x$-axis.$

$\noindent This implies (as mentioned before), that $P$ and $Q$ stay in the same `places', just with their names switched around. The circle is tangent to $P$ and $Q$ (because applying a reflection to the whole plane doesn't suddenly stop things from being tangent), which means that it is still tangent at $\left(5,7\tfrac{1}{2}\right)$ and $\left(5,-7\tfrac{1}{2}\right)$ in the reflected plane. This means the circle is still the same circle as before, so its centre is the same, so $(h,-k)$ must be the same as $(h,k)$, which means $k=-k\Rightarrow k = 0$.$

$\noindent Another way to see that $k$ must be 0 is that as the circle's centre is equidistant from $P$ and $Q$, it must lie on the perpendicular bisector of $P$ and $Q$, which means it lies on the $x$-axis. Hence $k=0$.$

erm i think this would make more sense to me if I the only thing I knew about ellipses was not that they were ovals ahahaha

edit: jks i understand the logic actually
idk how to sketch the ellipse though

InteGrand · Mar 25, 2016

Nailgun said:
erm i think this would make more sense to me if I the only thing I knew about ellipses was not that they were ovals ahahaha

edit: jks i understand the logic actually
idk how to sketch the ellipse though

Yeah, all the reflection thing is really saying is that if we flip the entire diagram of the ellipse about the horizontal axis, the picture remains the same because the ellipse is symmetric about the x-axis. Since the two points of interest are also reflections of each other about the x-axis, when we flip everything, the two points together are the same still, so the circle in the flipped case must be the same one as in the non-flipped case; and this'll happen precisely if the circle is centred on the x-axis (k = 0).

The second argument using the perpendicular bisector doesn't use anything about ellipses, it just uses the fact that the locus of points equidistant from two points in the plane is their perpendicular bisector, which is common knowledge for HSC 4U Complex Numbers locus questions.

An ellipse is basically a circle stretched/compressed in two perpendicular directions.

Nailgun · Mar 25, 2016

InteGrand said:
Yeah, all the reflection thing is really saying is that if we flip the entire diagram of the ellipse about the horizontal axis, the picture remains the same because the ellipse is symmetric about the x-axis. Since the two points of interest are also reflections of each other about the x-axis, when we flip everything, the two points together are the same still, so the circle in the flipped case must be the same one as in the non-flipped case; and this'll happen precisely if the circle is centred on the x-axis (k = 0).

The second argument using the perpendicular bisector doesn't use anything about ellipses, it just uses the fact that the locus of points equidistant from two points in the plane is their perpendicular bisector, which is common knowledge for HSC 4U Complex Numbers locus questions.

An ellipse is basically a circle stretched/compressed in two perpendicular directions.

Ah right right

then you have h=h and k=-k

or since their reflections of each other, they are the same distance from the x-axis, i.e their midpoint is 0

2016ers Chit-Chat Thread (4 Viewers)

Booty Connoisseur

Well-Known Member

Well-Known Member

Booty Connoisseur

Cole World

Well-Known Member

Cole World

Well-Known Member

Well-Known Member

Well-Known Member

Booty Connoisseur

Well-Known Member

Booty Connoisseur

Well-Known Member

Cole World

Booty Connoisseur

Well-Known Member

Cole World

Well-Known Member

Cole World

Users Who Are Viewing This Thread (Users: 0, Guests: 4)