A curve is defined by the parametric equations x = acos^2 t, y = asin^3 t
for range 0<= t <= pi/2 where a is a positive constant.
(i) Find and simplify an expresion for dy/dx in terms of t
(ii) Find the equation of the noraml at the point on the curve where
t = pi/6
(ii) If the normal cuts the the axis at point M and N, find the area of the triangle of the triangle OMN, in terms of a, where O is the origin.
OK.
(i)after finding dx/dt and dy/dt I simplfy to get -3/2(sin t)
(ii) The grad. of tangent is -3/4 when t = pi/6. Therefore the normal will be 4/3
(iii) I'm lost with this part...
Am I correct up to (iii) and can anyone help???
Thanks
The equation of the normal will be , where is the point on the curve given by , in other words the point .
Once you know the equation for the normal, you should be able to find the points M and N where it crosses the axes. That will tell you the lengths OM and ON, and then you can calculate the area of the triangle OMN as .
Thank you both for your comments. I've attempted this again and wondered if you could check it for me. I'm also confused about the "a" the positive constant, as I dont know this I am right to expect it in my answer?
Here is what I've done:
x = acos^2 t, y = asin^3 t
... so when t = pi/6
x = 3a/4
y = a/8
Putting these in the equation of the line:
y - (a/8) = 4/3(x - (3a/4))
24y = 32x - 21a
With some manipulation when x, y are set to zero the intercepts can be found:
x = 0 then y = -7a/8 (call this point M)
y = 0 then x = 21a/32 (call this point N)
So the lenghts OM = 7a/8 and ON = 21a/32
Therefore the area is 1/2*(7a/8 * 21a/32) which is (147a^2)/512.
Is this right?
Thanks, D
x = 0 the y intercept