Vastor Posted May 20, 2011 Share Posted May 20, 2011 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? Link to comment Share on other sites More sharing options...
DrRocket Posted May 20, 2011 Share Posted May 20, 2011 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. 1 Link to comment Share on other sites More sharing options...
randomc Posted May 20, 2011 Share Posted May 20, 2011 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. 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 20, 2011 Author Share Posted May 20, 2011 so, here the graph, so p > 21 ? and any calculation to get the result? Link to comment Share on other sites More sharing options...
DrRocket Posted May 20, 2011 Share Posted May 20, 2011 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. 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 20, 2011 Author Share Posted May 20, 2011 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? Link to comment Share on other sites More sharing options...
DJBruce Posted May 20, 2011 Share Posted May 20, 2011 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 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 20, 2011 Author Share Posted May 20, 2011 [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? Link to comment Share on other sites More sharing options...
imatfaal Posted May 20, 2011 Share Posted May 20, 2011 (edited) 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 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 20, 2011 Author Share Posted May 20, 2011 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? Link to comment Share on other sites More sharing options...
imatfaal Posted May 20, 2011 Share Posted May 20, 2011 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. 1 Link to comment Share on other sites More sharing options...
randomc Posted May 21, 2011 Share Posted May 21, 2011 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. 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 21, 2011 Author Share Posted May 21, 2011 (edited) [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 Link to comment Share on other sites More sharing options...
randomc Posted May 21, 2011 Share Posted May 21, 2011 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. 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 21, 2011 Author Share Posted May 21, 2011 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.. Link to comment Share on other sites More sharing options...
Vastor Posted May 21, 2011 Author Share Posted May 21, 2011 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 Link to comment Share on other sites More sharing options...
Vastor Posted May 22, 2011 Author Share Posted May 22, 2011 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 ??? Link to comment Share on other sites More sharing options...
the tree Posted May 22, 2011 Share Posted May 22, 2011 (edited) 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 1 Link to comment Share on other sites More sharing options...
Vastor Posted May 24, 2011 Author Share Posted May 24, 2011 so p> plus minus 10? so, from the grap p > 10 and p < -10 so, how come p > plusminus 10 = p < -10?? Link to comment Share on other sites More sharing options...
Vastor Posted May 25, 2011 Author Share Posted May 25, 2011 almost 1 day already and i need the answer or at least some clue :/ Link to comment Share on other sites More sharing options...
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