A curve C is given by the parametric equations
x=t^2+3 & y=t(t^2+3)
i) Find the Cartesian equation of C and hence show, using an algebraic method, that the
curve is symmetrical about the x-axis.
ii) Show, using an algebraic method, that there is no part of the curve for which x <3.
iii) Sketch the curve and label the axial intercepts.
Guys, i totally dont know how to go about solving this. I tried to make (t^2+3) the subject of the equation in order to equate x and y but i end up with the "t" for the y part ie y/t.
The curve C has the equation y =(x^2+2x) /(x+a)
, where a is a constant such that a =/= 0 and a =/= 2 .
(i) Find the equations of the asymptotes of C.
(ii) Find dy/dx. Hence find the set of values of a for which C has no stationary points.
(iii) Sketch C for a = −1, showing clearly the axial intercepts, asymptotes and stationary
points (if any).
Hence, or otherwise, find the range of values of k such that the equation
2 ( 1) 3 2 x + x = kx x − has exactly one real root.
I cant solve for (ii) after i obtain dy/dx to be 1 - (a^2-2a)/(x+a)^2. So what if C has no stationary points? I tried equating dy/dx to 0 but it felt wrong so...
10 The curve
C is defined by the equation y = x/(x^2-5x+4)
C
, indicating clearly the axial intercept(s), asymptote(s) and turning
point(s).
k such that the equation x/(x^2-5x+4) = k/x
has exactly 2 real roots.
stuck at (iii). i noticed that x was the common term on both sides of the equation so i divided both sides by x. (am i correct by doing this?) Then after manipulating, i got kx^2-5kx+4k-1=0. Next, i use b^2-4ac but my answers for k are 0 and -9/4 which are wrong as k should be positive. Hmm...
In general, i find that i cannot solve the question when it involves the curve having no stationary points. Any tips on this? Thanks everyone :D:D
Opps, kinda messy. Here.
A curve C is given by the parametric equations
x=t^2+3 & y=t(t^2+3)
i) Find the Cartesian equation of C and hence show, using an algebraic method, that the
curve is symmetrical about the x-axis.
ii) Show, using an algebraic method, that there is no part of the curve for which x <3.
iii) Sketch the curve and label the axial intercepts.
Guys, i totally dont know how to go about solving this. I tried to make (t^2+3) the subject of the equation in order to equate x and y but i end up with the "t" for the y part ie y/t.
______________________________________________________________________
The curve C has the equation y =(x^2+2x) /(x+a)
, where a is a constant such that a =/= 0 and a =/= 2 .
(i) Find the equations of the asymptotes of C.
(ii) Find dy/dx. Hence find the set of values of a for which C has no stationary points.
(iii) Sketch C for a = −1, showing clearly the axial intercepts, asymptotes and stationary
points (if any).
Hence, or otherwise, find the range of values of k such that the equation
2 ( 1) 3 2 x + x = kx x − has exactly one real root.
I cant solve for (ii) after i obtain dy/dx to be 1 - (a^2-2a)/(x+a)^2. So what if C has no stationary points? I tried equating dy/dx to 0 but it felt wrong so...
_________________________________________________________________
10 The curve
C is defined by the equation y = x/(x^2-5x+4)
Find a positive value for k such that the equation x/(x^2-5x+4) = k/x
has exactly 2 real roots.
stuck at (iii). i noticed that x was the common term on both sides of the equation so i divided both sides by x. (am i correct by doing this?) Then after manipulating, i got kx^2-5kx+4k-1=0. Next, i use b^2-4ac but my answers for k are 0 and -9/4 which are wrong as k should be positive. Hmm...
In general, i find that i cannot solve the question when it involves the curve having no stationary points. Any tips on this? Thanks everyone :D:D
Hi,
Question:
A curve C is given by the parametric equations x=t^2+3 & y=t(t^2+3).
(i) Find the Cartesian equation of C and hence show, using an algebraic method, that the curve is symmetrical about the x-axis.
Approach:
y = tx, so t = y/x. Therefore x = (y/x)^2 + 3...
To show that C is symmetrical about x-axis, replace y by -y and you'll still obtain the same expression.
Thanks!
Cheers,
Wen Shih
Hi,
Question:
The curve C has the equation y =(x^2+2x) /(x+a), where a is a constant such that a =/= 0 and a =/= 2.
(ii) Find dy/dx. Hence find the set of values of a for which C has no stationary points.
I can't solve for (ii) after i obtain dy/dx to be 1 - (a^2-2a)/(x+a)^2. So what if C has no stationary points? I tried equating dy/dx to 0 but it felt wrong so...
Approach:
dy/dx = 1 - (a^2 - 2a)/(x + a)^2 = [ (x + a)^2 - a^2 - 2a ] / (x + 1)^2.
Since C has no stationary points, dy/dx is never zero, so that the numerator expression is never zero (i.e. no real roots).
The numerator expression becomes x^2 + 2ax - 2a, after simplification.
Consider discriminant < 0 to find the set of values of a.
Thanks!
Cheers,
Wen Shih
Hi,
Question:
The curve C is defined by the equation y = x/(x^2-5x+4).
Find a positive value for k such that the equation x/(x^2-5x+4) = kx has exactly 2 real roots.
Approach:
Sketch the graph of y = x/(x^2-5x+4).
One should realise that y = kx (where k > 0) will intersect y = x/(x^2-5x+4), provided that y = kx is tangent to the curve at the origin.
Find the value of dy/dx at the origin, which will be the value of k.
Using the algebraic method would only give k = -4/9.
Thanks for sharing a nice question in which the graphical method works better than the algebraic one!
Cheers,
Wen Shih
P.S. To generalise the problem, y = 1/4 x and y = -4/9 x each intersect the curve twice. Take a look at this diagram:
http://www.freewebs.com/weews/Graph%20(13%20Aug%2009).pdf
Woops, for the last question, x/(x^2-5x+4) = kx instead of x/(x^2-5x+4) = k/x -.-
Woops, for the last question, x/(x^2-5x+4) = kx instead of x/(x^2-5x+4) = k/x -.-
Hi,
I've changed my solution in response to your change of question. Thanks!
Cheers,
Wen Shih