conics

niharika_kumar · 2015-03-25 18:27:15

The tangent and normal at the point P(at²,2at) to the parabola y²=4ax meet the x-axis in T and G respectively, then find the angle at which the tangent at P to the parabola is inclined to tangent at p to the circle through P,T,G.

Olinguito · 2015-03-25 21:26:41

The gradient at P to the curve is straightforward to compute: it's . Also T is the point and P is .

Since TPG is a right angle, TG is the diameter of the circle passing through P, T, G. Therefore the centre of the circle lies on the x-axis; it is in fact . Thus the equation of the circle is

[list=*]
[*]

[/*]
[/list]

Implicit differentiation gives

[list=*]
[*]

[/*]
[/list]

at P.

Hence the angle between the two tangents is

[list=*]
[*]

[/*]
[/list]

PS: It's actually

[list=*]
[*]

[/*]
[/list]

since the gradient to the curve will always be steeper than the gradient to the circle.

Last edited by Olinguito (2015-03-25 21:32:05)

Bob · 2015-03-25 22:28:17

hi

Like that solution. Then you can use

to simplfy this

Bob

Olinguito · 2015-03-25 22:57:17

Oh! Thank you Bob, I forgot about the tan formula.

Bob · 2015-03-26 05:17:38

I think there's a geometric solution too, but I've been busy today in the garden. I'll have a go now.

LATER EDIT:

AS TPG = 90, => TG is a diameter. Let C be the centre of the circle, and let FPH be the tangent to the circle at P (F on the 'T' side of the circle)

FPC = 90

x = FPT = CPG and y = TPC = GPH = PTC.

WE know that TP has gradient 1/t => tan(y) = 1/t and, as x + y = 90, => tan(x) = -t

Bob

niharika_kumar · 2015-03-31 15:33:35

thank you for the detailed solution.

niharika_kumar · 2015-04-05 23:09:38

Let

be squares such that for each n≥ 1, the length of a side of equals the length of a diagonal of .If the length of the side of is 10 cm, then for which of the following values of n is the area of less than 1 sq. cm.

Bob · 2015-04-06 00:26:01

hi Niharika

Hopefully this diagram will help:

If ABCD is S(n+1) then DECF is S(n).

The area of S(1) is 10 x 10 = 100.

DE = 5√ 2, so S(2) is 25 x 2 = 50.

So just continue this sequence.

Bob

Agnishom · 2015-04-06 02:53:24

niharika_kumar · 2015-04-10 03:33:10

oh thanks.

Well, I would be happy if you will help me by suggesting some ways on how to have a good command in conics. Pls help.

Agnishom · 2015-04-10 03:37:03

What has this question got to do with conics?

Bob · 2015-04-10 03:54:04

hi Niharika,

Do you mean cones / pyramids / that sort of solid or, rather, conic sections like circle, ellipse, parabola, hyperbola?

Bob

Agnishom · 2015-04-10 04:19:45

She means conics

Bob · 2015-04-10 05:21:22

Look here first:

https://www.mathsisfun.com/geometry/conic-sections.html

then come back to the forum with a more specific query.

Bob

niharika_kumar · 2015-04-10 17:21:19

I meant conic sections.
my query was that how can I analyze the problems based on conic sections efficiently?

Bob · 2015-04-10 21:18:21

hi Niharika,

Imagine two identical cones one up-side-down with their vertices touching. The cones extend to infinity. Assume they are hollow.

Any plane will cut through the cone(s) and the resulting 2D shape is called a conic section.

There are five possibilities:

A circle: this occurs when the cut is horizontal.

An ellipse: this occurs when the cut is sloping at an angle less than the angle of the cone.

A parabola: this occurs when the cut is exactly parallel with the angle of the cone.

A hyperbola: this occurs when the cut is sloping at an angle greater than the angle of the cone. The section will be in two parts; one from the top cone; and one from the up-side-down cone.

Two straight lines: this occurs when the cut is vertical through the common vertex.

The general formula for any conic section takes the form:

[This arises from the general equation for the cones intersecting with the equation for a plane in 3D.]

If E = 0 and A = B the conic will be a circle.

If A ≠ B but they have the same sign, then the conic will be an ellipse.

If either A or B = 0 (but not both) then the conic will be a parabola.

If A and B have opposite signs, then the conic will be a hyperbola.

If the equation is factorable as

then the conic is two straight lines.

The ellipse, hyperbola and two straight lines have two axes of symmetry. The parabola has one. The circle has many.

But the issue is complicated because the format of the equations can be changed by a change of axes.

eg. With the format above, a hyperbola will have two axes of symmetry and these might be the x and y axes by suitable choice of the coefficients. By changing the axes so that the asymptotes are the axes, the equation becomes:

Does that help?

Bob

Agnishom · 2015-04-10 21:52:22

Probably not. There can be harder conic problems

niharika_kumar · 2015-04-11 01:08:02

thanks bob as you really tried hard.But agnishom is right.I certainly know the basics but get stuck when the problems demand applications of many properties.One of the examples of such problem was the first one that I asked in this thread.

Agnishom · 2015-04-11 01:28:00

Conic Sections is probably the hardest chapter in +2 level. I'll probably be able to master integration techniques with enough hardwork.

Last edited by Agnishom (2015-04-11 01:33:13)

Bob · 2015-04-11 02:58:44

hi Niharika,

Oh!

What I'm lacking here is information. Here's a suggestion:

Post three problems: (1) One you have done and got right; (2) One you are struggling to complete (say what you have tried); and (3) One you don't know where to start.

Bob

niharika_kumar · 2015-04-13 15:26:52

the question I easily made:-
#3 normals to a parabola y^2 =4ax meet at (h,k).Then show that h>4a.

The question I have tried:-
#Find the locus of point such that 2 of the 3 normals drawn from them to the parabola coincide.
This is what i tried;

Now what to do after this.

This is what I don't know where to start.
#Find the locus of the vertices of the family of parabolas

Bob · 2015-04-13 23:27:29

hi Niharika,

When I get set a question, I assume that the questioner has devised one that does have a sensible solution. If you're not told much that should mean that if you do whatever is possible, it ought to lead to an answer. So these questions may look daunting but let's see what is possible and hope my faith is upheld.

Q2. I had to remind myself what the equation of a normal is, based on a parameter m. My version isn't quite the same as yours. So I looked it up in a book and mine is correct. I think you had the gradient of a normal as m. It should be -m.

So your starting point was good but needs a small adjustment.

If you sketch the parabola, you'll see that typical normals have negative gradients.

So if you're given x and y, you get a cubic in m, which you'd expect to have three solutions ie values of m.

I couldn't see that it would be very easy to factorise the cubic, especially with x and y in it, so I thought maybe I could start from the other end and assume the cubic has the form:

That has two coincident roots (p twice) and one separate one (q), so it fits the question.

You can probably do the rest yourself. Expand the cubic and compare coefficients with (1).

The algebra isn't that nice but, remember: "it ought to lead to an answer"

So eliminate p and q and you'll be left with an equation containing x, y and a. That will be the required locus.

Q3. I had no idea where the vertex would be for that equation. I know you cannot do this in a test but it's worth having a look at the graph. You'll find it here:

http://www.mathsisfun.com/data/function … +a^2x/2-2a

The function grapher has a slider that lets you alter the value of 'a', so you'll be able to see how the vertex changes for yourself. Of course, that's no answer to the question, but hopefully it will make it seem more manageable.

If the graph was simpler (eg. y = x^2) then you would know all about the vertex, so what I did next was to change the coordinates to bring the vertex to the origin. This is a handy technique that may help with other questions. You try to make the equation simpler by changing to a new x and y coordinate system. Here's my method:

One simplification that suggested itself to me was to change the x coordinate to X = ax. the equation then becomes:

Already, it doesn't look so frightening. Now if I 'complete the square' for that quadratic I can set it up for the next coordinate change.

So now I can make a second change of coordinates thus:

ie.

Now that parabola is very simple. We know it has a vertex at (0,0) for all a.

So put

and that's nearly done. You want the locus of 'a' so eliminate it from the above and the resulting equation in the old y and x will be the locus.

LATER EDIT: and you can put this locus into the function grapher as the second function to see if this works. (I found I needed to change a to -a to get the second half of the solution. I didn't find a way to make a negative on the slider.)

I suggest you have a try at these to finish them off and then see if you can do another yourself. If not, you know where to come.

Bob

Last edited by Bob (2015-04-15 19:03:27)

niharika_kumar · 2015-04-17 03:16:36

thanks bob.

and yes I will surely be disturbing you with other problems on conic section as well.:)

Math Is Fun Forum

#1 2015-03-25 18:27:15

conics

#2 2015-03-25 21:26:41

Re: conics

#3 2015-03-25 22:28:17

Re: conics

#4 2015-03-25 22:57:17

Re: conics

#5 2015-03-26 05:17:38

Re: conics

#6 2015-03-31 15:33:35

Re: conics

#7 2015-04-05 23:09:38

Re: conics

#8 2015-04-06 00:26:01

Re: conics

#9 2015-04-06 02:53:24

Re: conics

#10 2015-04-10 03:33:10

Re: conics

#11 2015-04-10 03:37:03

Re: conics

#12 2015-04-10 03:54:04

Re: conics

#13 2015-04-10 04:19:45

Re: conics

#14 2015-04-10 05:21:22

Re: conics

#15 2015-04-10 17:21:19

Re: conics

#16 2015-04-10 21:18:21

Re: conics

#17 2015-04-10 21:52:22

Re: conics

#18 2015-04-11 01:08:02

Re: conics

#19 2015-04-11 01:28:00

Re: conics

#20 2015-04-11 02:58:44

Re: conics

#21 2015-04-13 15:26:52

Re: conics

#22 2015-04-13 23:27:29

Re: conics

#23 2015-04-17 03:16:36

Re: conics

Board footer