# equation of tangentsWatch

This discussion is closed.
#1
what is tangent to (x^2/a^2)+(y^2/b^2)=1
at [a cos(theta), b sin(theta)]
0
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.
0
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
0
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
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
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.
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