equation of tangents Watch

This discussion is closed.
ronnie
Badges: 0
#1
Report Thread starter 13 years ago
#1
what is tangent to (x^2/a^2)+(y^2/b^2)=1
at [a cos(theta), b sin(theta)]
0
steve2005
Badges: 17
Rep:
?
#2
Report 13 years ago
#2
(Original post by ronnie)
what is tangent to (x^2/a^2)+(y^2/b^2)=1
at [a cos(theta), b sin(theta)]
I don't think the co-ordinates specified lie on the elipse. See attached
picture.
Attached files
0
steve2005
Badges: 17
Rep:
?
#3
Report 13 years ago
#3
(Original post by steve2005)
I don't think the co-ordinates specified lie on the elipse. See attached
picture.
I have calculated the tangent at x=2 for an elipse with a=3, b=4. It is necessary to calculate the value of y and then to plug these into the derivative.

See attached elipse with tangent drawn and the calculation. I have not used
theta as I think the original question is flawed. ( I might be wrong and would
welcome any comments) Sorry picture isn't larger but I can limit on uploading is 100 kb
Attached files
0
JamesF
Badges: 1
Rep:
?
#4
Report 13 years ago
#4
(acosθ, bsinθ ) are the co-ordinates of the points on the ellipse. To see this, substitute them in to the LHS of the equation and you get
(x^2/a^2)+(y^2/b^2) = sin²θ + cos²θ = 1.

Anyway, to find the tangent, differentiate to find the gradient,

(2x/a²) + (2y/b²).(dy/dx) = 0
(2x/a²) = -(2y/b²).(dy/dx)
xb²/(ya²) = dy/dx

Sub in the co-ordinates
dy/dx = -bcotθ/a

Then use y-y0 = bcotθ/a .(x - x0)
0
steve2005
Badges: 17
Rep:
?
#5
Report 13 years ago
#5
(Original post by JamesF)
(acos?, bsin? ) are the co-ordinates of the points on the ellipse. To see this, substitute them in to the LHS of the equation and you get
(x^2/a^2)+(y^2/b^2) = sin²? + cos²? = 1.

Anyway, to find the tangent, differentiate to find the gradient,

(2x/a²) + (2y/b²).(dy/dx) = 0
(2x/a²) = -(2y/b²).(dy/dx)
xb²/(ya²) = dy/dx

Sub in the co-ordinates
dy/dx = -bcot?/a

Then use y-y0 = bcot?/a .(x - x0)
Your solution looks good BUT I can't see how the points on the ellipse can
have co-ordinates (acos?, bsin? ). Can you please explain further?

I can see my mistake. I forgot to square.
0
JohnSPals
Badges: 0
Rep:
?
#6
Report 13 years ago
#6
(Original post by ronnie)
what is tangent to (x^2/a^2)+(y^2/b^2)=1
at [a cos(theta), b sin(theta)]
You'll almost certanly get asked a question like this in your P5/FP2 exam. If not for the ellipse, then for the hyperbola or parabola (it was the parabola this year). We went through the proofs in class, but I'm sure there'll be a proof in the Heinemann textbook.
0
X
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the Arts London
    MA Performance Design and Practice Open Day Postgraduate
    Thu, 24 Jan '19
  • Coventry University
    Undergraduate Open Day Undergraduate
    Sat, 26 Jan '19
  • Brunel University London
    Undergraduate Experience Days Undergraduate
    Sat, 26 Jan '19

Are you chained to your phone?

Yes (78)
19.4%
Yes, but I'm trying to cut back (163)
40.55%
Nope, not that interesting (161)
40.05%

Watched Threads

View All