straight line and curve(math)

[Vastor]

the Question say,

 

Find the range of value of p for which the straight line [math]y = 2x + p[/math] does not intersect the curve [math]x^2 + y^2 = 20 [/math]

 

so, i dun have any idea how to start..

 

the textbook don't tell me anything..

 

so, help anyone?

[DrRocket]

the Question say,

 

Find the range of value of p for which the straight line [math]y = 2x + p[/math] does not intersect the curve [math]x^2 + y^2 = 20 [/math]

 

so, i dun have any idea how to start..

 

the textbook don't tell me anything..

 

so, help anyone?

 

 

Draw yourself a graph of

 

[math]x^2 + y^2 = 20 [/math]

 

 

Also graph [math]y = 2x + p[/math] on he same axes for several values of p. An approach should then suggest itself.

[randomc]

The line perpendicular to y=2x+p with end-points on the circle and at the origin fixes the y-intercept where you need it.

[Vastor]

 

 

so, here the graph, so p > 21 ?

 

and any calculation to get the result?

 

[DrRocket]

 

 

so, here the graph, so p > 21 ?

 

and any calculation to get the result?

 

 

Try again. [math] x^2+y^2 = 20[/math] does not describe a parabola.

[Vastor]

Try again. [math] x^2+y^2 = 20[/math] does not describe a parabola.

 

i'm not get this, the question is in chapter of Quadratic Function, and Q.F. without parabola?

[DJBruce]

i'm not get this, the question is in chapter of Quadratic Function, and Q.F. without parabola?

 

[math]x^{2}+y^{2}=20[/math] is a quadratic form, ie: a polynomial of degree 2 in multiply variables. So it does fit in a chapter about quadratics.

 

As for what this graph should look like:

 

http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2%3D20

[Vastor]

[math]x^{2}+y^{2}=20[/math] is a quadratic form, ie: a polynomial of degree 2 in multiply variables. So it does fit in a chapter about quadratics.

 

As for what this graph should look like:

 

http://www.wolframal...E2%2By%5E2%3D20

 

and the 'unlucky' part about this is, i never learn this type of graph, so, there's no way i can plot it by using my knowledge. but, wth this came as my question -,-

 

or maybe there any alternative way, a type of calculation or formula where i can get the answer without knowing what the graph actually about?

[imatfaal]

maths teachers like to give you questions just outside your experience. plot a few graphs and think about what doc rock, djb and random said above.

Think back to geometry - what is a line that just touches a circle? - and what angle is it to the radius? then go back to random's post and draw another graph

once you have twigged - its just algebra

Edited May 20, 2011 by imatfaal

[Vastor]

maths teachers like to give you questions just outside your experience. plot a few graphs and think about what doc rock, djb and random said above.

Think back to geometry - what is a line that just touches a circle? - and what angle is it to the radius? then go back to random's post and draw another graph

once you have twigged - its just algebra

 

ermm, can't think of any line I ever learn that 'just touch' the circle

 

so, i google and found... tangent?, and i only know about tan because it's appear on calculator (not really know about it, what it's use for etc)

 

"what angle is it to the radius?" well, i hope i know what is that and how to calculate that

 

after think so hard, the only thing that i can expect is all of you trying to say that y = p +2x is tangent to the circle?

[imatfaal]

Look at this page Tangent - you don't need to calculate angle of tangent to curve. You need to either break out the graphing and understand the pictures, or delve into calculus and algebra.

[randomc]

You're trying to find the coordinates of the point at which y=2x+p is tangent to the curve. To find points of intersection of a line and curve, just substitute the equation of the line into the curve (or circle, whatever).

 

y=mx+p

x^2 + y^2 = r^2

 

gives something of the form

 

x^2 + (mx+p)^2 = r^2

 

but since in this problem you get

 

x^2 + (2x + p)^2 = 20

 

its difficult to sort out all the variables.

 

You can still solve it by the same approach of substitution, but instead by using the line that passes through the origin that is perpendicular to y=2x+p. Since the slopes of two perpendicular lines are the negative reciprocal of each other, the equation of this perpendicular line is

 

y = -1/2x ( +0, because it passes through the origin)

 

This line joins the centre of the circle to a point on the circle, and because it is perpendicular to y=2x+p, the point at which it intersects the circle is the same point at which y=2x+p is tangent. So substituting y= -1/2x into the equation of the circle gives the coordinates of a point on y = 2x+p.

 

x^2 + (-1/2x)^2 = 20

(solve for x here then y in x^2 + y^2=20)

 

Then you can rewrite y=2x+p in point-slope from which you can find the range you are asked for.

[Vastor]

[math] y = -\frac{1}{2}x + 0[/math]

 

[math] x^2 + (- \frac{1}{2}x)^2 = 20[/math]

 

[math] x^2 + \frac{1}{4}x^2 = 20[/math]

 

[math] (1 + \frac{1}{4}) x^2 = 20[/math]

 

[math] (\frac{4}{4} + \frac{1}{4}) x^2 = 20[/math]

 

[math] (\frac{5}{4}) x^2 = 20[/math]

 

[math] x^2 = 16[/math]

 

[math] x = 4[/math]

 

[math] x^2 + y^2 = 20 [/math]

 

[math] (4)^2 + y^2 = 20 [/math]

 

[math] 16 + y^2 = 20 [/math]

 

[math] y^2 = 4 [/math]

 

 

[math] y = 2 [/math]

 

and, the point-slope form is,

 

 

[math] y - 2 = 2(x - 4)[/math]

 

[math] y = 2x - 6 [/math]

 

[math] y = 2x + p[/math]

 

so, [math] p = -6[/math] ???

Edited May 21, 2011 by Vastor

[randomc]

x^2=16

x = plus or minus 4

 

Anyway, I think i'm just distracting you from what you should have been doing, which is looking at the graph.

[Vastor]

x^2=16

x = plus or minus 4

 

Anyway, I think i'm just distracting you from what you should have been doing, which is looking at the graph.

 

and [math]y = +/- 2[/math]

 

which would result to

 

[math] p = +/- 6[/math]

 

and

 

[math]p = +/- 10[/math]

 

ok, now what? I'm lost..

[Vastor]

ha3, it's seems I fail to understand the concept that you guys try to tell me, well, nevermind, tomorrow, I can ask my teacher. ^^

 

anyway, thanks all

 

 

 

[Vastor]

and this is what she said...

[math]y = 2x + p[/math]

 

[math]x^2 + y^2 = 20[/math]

 

 

 

[math]x^2 + (2x+p)^2 = 20[/math]

 

[math]x^2 + 4x^2 + 4px + p^2 = 20[/math]

 

[math]5x^2 + 4px + p^2 -20 = 0[/math]

 

 

then, used discriminant [math] b^2 - 4ac < 0[/math] because it doesn't intersect

 

[math](4p)^2 -4(5)(p^2-20) < 0[/math]

 

[math]16p^2 -20p^2 + 400 < 0[/math]

 

[math]-4p^2 < -400[/math]

 

[math]p^2 < 100[/math]

 

[math]p < 10[/math]

and this is my first calculation(or second) for this question , i ask here because p < 10 would make the graph intersect doesn't ???

[the tree]

Yeah that definitely intersects (see graph), but at [imath]p=10[/imath], (graph) the line is clearly tangent to the curve.

 

Where you majorly went wrong:

[math]-4p^2 < -400[/math]

[math]p^2 < 100[/math]

Check your knowledge of solving inequalities.

 

 

[math]p^2 < 100[/math]

[math]p < 10[/math]

As welll as the inequality being the wrong way round, this is wrong in another way. Which you should be able to figure out by, y'know, looking at the graph. Edited May 22, 2011 by the tree

[Vastor]

so p> plus minus 10?

 

so, from the grap p > 10 and p < -10

 

so, how come p > plusminus 10 = p < -10??

[Vastor]

almost 1 day already and i need the answer or at least some clue :/