# Conic sections

• January 28th 2008, 05:59 AM
Tangera
Conic sections
Hello! I wonder if there's a better way of solving the questions below rather than sketching out the curves and circles or substiting in values of r and checking by trial and error. Please give suggestions. Thank you! :)

1) Consider the graph described by the equation y = tan x. The equation of a circle centred at the origin may also be written as
y = + (r^2 - x^2)^(1/2)

after rearrangement. Give an equation of a circle centred at the origin that

(a) intersects the graph at exactly 2 points

--> Based on my understanding of the tangent graph,
y = + (pi^2 - x^2)^(1/2) fulfils the requirement
:)

(b) intersects the graph at exactly 10 points

2) Is there a circle centred at the origin that interesects the graph at exactly 4 points? If so, what is its equation? If not, why not?

3) Can a circle centred at the origin intersect the graph at an odd number of points?
--> I don't think so, because when the circle cuts the graph at one point,say, point A, it will also cut another point on the graph that is symmetrical to point A about the origin. [Nature of the y = tan x graph]

Thank you for helping me!
• January 28th 2008, 07:13 AM
TKHunny
a) pi is no good. That's six intersections. You must keep it under 2.51 or so. It would be a nice challenge to get exactly 4 intersections.

b) Try just a little over 5.6. Really, you just have to figure out how to solve using these equations. I'm kind of wondering why the tangent is in a "conic section" problem. Again, finding exactly 8 intersections would be a good problem.

c) It would help me if you specified the exact "nature" of the tangent. Specifically, tan(x) = -tan(-x). This property has a name. What is it? In other words, if you hit y = tan(x) at (a,b), you MUST hit it again at (-a,-b), a reflection across the Origin.
• January 28th 2008, 06:05 PM
mr fantastic
Quote:

Originally Posted by Tangera
Hello! I wonder if there's a better way of solving the questions below rather than sketching out the curves and circles or substiting in values of r and checking by trial and error. Please give suggestions. Thank you! :)

1) Consider the graph described by the equation y = tan x. The equation of a circle centred at the origin may also be written as
y = + (r^2 - x^2)^(1/2)

after rearrangement. Give an equation of a circle centred at the origin that

(a) intersects the graph at exactly 2 points
--> Based on my understanding of the tangent graph,
y = + (pi^2 - x^2)^(1/2) fulfils the requirement
:) Mr F says: That won't work. Play around with $0 < r < \frac{\pi}{2}$ ....

(b) intersects the graph at exactly 10 points Mr F says: Play around with $\frac{3 \pi}{2} < r < \frac{5 \pi}{2}$ ....

2) Is there a circle centred at the origin that interesects the graph at exactly 4 points? If so, what is its equation? If not, why not? Mr F says: Nice question .... as I played around with some rough graphs it certainly looks possible ...... You obviously need the circle to be TANGENT (touching but not crossing) to the parts of the tan graph between
$\frac{\pi}{2}$ and $\frac{3\pi}{2}$,
and
$\frac{-\pi}{2}$ and $\frac{-3\pi}{2}$ .....

And, as TKH pointed out "if you [touch] y = tan(x) at (a,b), you MUST [touch] it again at (-a,-b)" .... If you think about it, you'll see that the radius of the circle will $\sqrt{a^2 + b^2}$, so the question boils down to getting a and b ..... See my main reply below.

3) Can a circle centred at the origin intersect the graph at an odd number of points?
--> I don't think so, because when the circle cuts the graph at one point,say, point A, it will also cut another point on the graph that is symmetrical to point A about the origin. [Nature of the y = tan x graph] Mr F says: You're right. It gets back to what TKH said about (a, b) and (-a, b) ........ You will always have PAIRS of intersection points => only an even number.

Thank you for helping me!

2) Hopefully you can see that the line passing through the two points of the circle tangent to the tan graph has the form $y = -\frac{b}{a} x$ .....

So you require

$\tan x = -\frac{b}{a} x$

But x = a is a solution. So

$\tan a = -b$ .... (1)

Now note that the tangent to a circle at any point is always perpendicular to the radius of the circle. So the tangent to the circle at the touching points has to be perpendicular to $y = -\frac{b}{a} x$. The gradient of $y = -\frac{b}{a} x$ is $-\frac{b}{a}$. Therefore the gradient of the tangent at the touching point has to be $\frac{a}{b}$. Therefore the gradient of the tan graph at the touching point has to be $\frac{a}{b}$.

But the gradient of y = tan x is given by $\frac{1}{\cos^2 x}$. So you also require

$\frac{1}{\cos^2 a} = \frac{a}{b} \Rightarrow \cos^2 a = \frac{b}{a}$ .... (2)

Now solve (1) and (2) simultaneously: Sub (1) into (2) and re-arrange:

$a \cos^2 a = -\tan a$

This is a transcendental equation and cannot be solved algebraically. Only approximate solutions can be got (using a graphics or CAS calculator) ...... I get the largest negative solution to be approximately a = -2.31981 (radians). Therefore b = 1.07555.

So the radius of the circle will be approximately $r = \sqrt{a^2 + b^2} = 2.55702$

It all seems to check out when I draw the graphs.

I realise there's a lot to digest here. Take your time - any questions, make sure you ask.
• January 31st 2008, 03:17 AM
Tangera
Aieee. I have many questions...(Doh) :confused:

1a) Ah! I think it was a careless mistake on my part. Thank you (tkhunny and mr fantastic) for correcting me!

b) Ermm... may I know how did both of you come to an answer? Was it by trial and error?

2) I understand that I need to find a tangent to the y = tan x graph for
(pi/2) < x < (3pi/2), but why is the general equation of passing through the two points of the circle tangent to the tan graph y= (-b/a)x? Why does it have a negative gradient? If the points that the tangent touch are (a,b) and (-a, -b), shouldn't the gradient be positive? Just to confirm, there is no constant term in the equation because the line passes through the origin so the origin is the y-intercept right? :confused:

Just to confirm again, the general rule for the radius is because of Pythagoras' Theorem right?

What is a transcendental equation?

Why do you take the largest negative solution? Why not a positive solution?? :confused:

If R = 2.55702, the equation of the circle is: y = (2.55^2 - x^2)^(1/2). [R rounded off to 3 significant figures.]

Thank you! :)
• January 31st 2008, 04:11 AM
mr fantastic
I was probably a bit careless in my notation ..... I'll re-post my solution, making the necessary edits, corresponding to the required tangent points having coordinates (-a, b) and (a , -b). See the attachment for a graph of everything. The answers are still the same.

Quote:

Originally Posted by Tangera
Aieee. I have many questions...(Doh) :confused:

1a) Ah! I think it was a careless mistake on my part. Thank you (tkhunny and mr fantastic) for correcting me!

b) Ermm... may I know how did both of you come to an answer? Was it by trial and error? Mr F says: More or less (for myself, anyway) - just playing around drawing circles.

2) I understand that I need to find a tangent to the y = tan x graph for
(pi/2) < x < (3pi/2), but why is the general equation of passing through the two points of the circle tangent to the tan graph y= (-b/a)x? Why does it have a negative gradient? Mr F says: Look at the attachment and you'll clearly see why the gradient is negative.

If the points that the tangent touch are (a,b) and (-a, -b), shouldn't the gradient be positive? Mr F says: No. But the reason is not clear from my poor choice of notation for the points. Look at the attachment.

Just to confirm, there is no constant term in the equation because the line passes through the origin so the origin is the y-intercept right? :confused:
Mr F says: Correct. You can justify this either by appealing to symmetry, or deriving the equation from the two tangent points (-a, b) and (a, -b):

$m = \frac{-b - b}{a - (-a)} = -\frac{b}{a}$.

Then using the model $y - y_1 = m(x - x_1) \,$:

$y - (-b) = -\frac{b}{a} (x - a) \Rightarrow y + b = -\frac{b}{a}x + b \Rightarrow y = -\frac{b}{a}$.

Just to confirm again, the general rule for the radius is because of Pythagoras' Theorem right? Mr F says: Yes.

What is a transcendental equation? Mr F says: It's an equation involving 'mixed' functions eg. trig and polynomial, exponential and polynomial, trig and exponential etc.

Why do you take the largest negative solution? Why not a positive solution?? :confused: Mr F says: Gets back to my poor choice of notation ..... If you look at the attachment, you'll now see that the smallest postive value of a is required.

If R = 2.55702, the equation of the circle is: y = (2.55^2 - x^2)^(1/2). [R rounded off to 3 significant figures.] Mr F says: Almost. You need a $\pm$ in front .....

Thank you! :) Mr F says: You're welcome.

My next reply will be an edit of my first, corresponding to the required tangent points having coordinates (-a, b) and (a , -b). See the attachment for a graph of everything. The answers are still the same.

You could probably make the appropriate changes to the solution yourself.

By the way, I'm mildly surprised you didn't ask why the gradient of the tangent to y = tan x is given by $\frac{1}{\cos^2 x}$ ......
• January 31st 2008, 04:24 AM
mr fantastic
The following is an edited version of my first reply. This version takes the coordinates of the two tangent points to be (-a, b) and (a, -b) where a and b are positive numbers whose values need to be found.

Quote:

Originally Posted by mr fantastic
2) Hopefully you can see that the line passing through the two points of the circle tangent to the tan graph has the form $y = -\frac{b}{a} x$ .....

So you require

$\tan x = -\frac{b}{a} x$

But x = a is a solution (corresponding to the point (a, -b)) . So

$\tan a = -b$ .... (1)

Now note that the tangent to a circle at any point is always perpendicular to the radius of the circle. So the tangent to the circle at the touching points has to be perpendicular to $y = -\frac{b}{a} x$. The gradient of $y = -\frac{b}{a} x$ is $-\frac{b}{a}$. Therefore the gradient of the tangent at the touching point has to be $\frac{a}{b}$. Therefore the gradient of the tan graph at the touching point has to be $\frac{a}{b}$.

But the gradient of y = tan x is given by $\frac{1}{\cos^2 x}$. So at the point (a, -b), that is, x = a, you also require

$\frac{1}{\cos^2 a} = \frac{a}{b} \Rightarrow \cos^2 a = \frac{b}{a}$ .... (2)

Now solve (1) and (2) simultaneously: Sub (1) into (2) and re-arrange:

$a \cos^2 a = -\tan a$

This is a transcendental equation and cannot be solved algebraically. Only approximate solutions can be got (using a graphics or CAS calculator) ...... I get the smallest positive solution to be approximately a = 2.31981 (radians). Therefore b = 1.07555.

So the radius of the circle will be approximately $r = \sqrt{a^2 + b^2} = 2.55702$

It does check out when I draw the graphs.

I realise there might still be a lot to digest here. Take your time - any questions, make sure you ask.

That should clear things up a bit I hope.
• January 31st 2008, 04:53 AM
Tangera
Quote:

Originally Posted by mr fantastic
I was probably a bit careless in my notation ..... I'll re-post my solution, making the necessary edits, corresponding to the required tangent points having coordinates (-a, b) and (a , -b). See the attachment for a graph of everything. The answers are still the same.

My next reply will be an edit of my first, corresponding to the required tangent points having coordinates (-a, b) and (a , -b). See the attachment for a graph of everything. The answers are still the same.

You could probably make the appropriate changes to the solution yourself.

By the way, I'm mildly surprised you didn't ask why the gradient of the tangent to y = tan x is given by $\frac{1}{\cos^2 x}$ ......

Oh heee. the gradient of the tangent = dy/dx, and d (tan x)/dx is secant square x! *yay!*

I have another question(which contradicts what I asked previously, but now I am somewhat confused): Based on the graph, I think I understand why the smallest positive value of a is chosen, but why not the largest negative a value, since there is also a point (-a,b) that intersects the y = tan x curve?

Also, you mentioned I could find the general equation of the tangent by "appealing to symmetry". What is that? (Just curious, because I've never heard of it before.)

Thank you for the clarifications! :)
• January 31st 2008, 05:09 AM
mr fantastic
Quote:

Originally Posted by Tangera
Oh heee. the gradient of the tangent = dy/dx, and d (tan x)/dx is secant square x! *yay!* Mr F says: (Star)

I have another question(which contradicts what I asked previously, but now I am somewhat confused): Based on the graph, I think I understand why the smallest positive value of a is chosen, but why not the largest negative a value, since there is also a point (-a,b) that intersects the y = tan x curve?

Mr F says: Yes, you could do that. Well spotted. The largest negative value is -2.31981. This corresponds to the point (-a , b) and so you'd have -a = -2.31981 .....

Also, you mentioned I could find the general equation of the tangent by "appealing to symmetry". What is that? (Just curious, because I've never heard of it before.) Mr F says: I didn't quite say that. I said you could use symmetry to justify saying that the line that joins the tangent points also passes through the origin. Maybe best to forget I mentioned it - stick with the mathematical derivation of the equation of the line joining the two tangent points.

Thank you for the clarifications! :) Mr F says: You're welcome. Your question was both interesting and useful to me.

..