I am stuck on this question.
x = root 3 cos t, y = sin t (t\in [0,2pi])
point P is where t = 5pi/6, ie (-1.5,0.5)
The gradient of the tangent to the ellipse at the point with parameter t is -cos t/root 3 sin t, where t is not equal to 0, pi, 2 pi.
I need to find the equation of the tangent to the ellipse at the point P.
Can anyone help?
The gradient at point with parameter t is -cos t/root 3 sin t; and the point P has parameter t=5pi/6. So the gradient of the tangent will be -cos(5pi/6) / root (3 sin (5pi/6)) ie. 1/sqrt(2)
Again the co-ordinates of point P are (-1.5 , .5). So using Point Slope Form of equation of straight line, the tangent's equation will be (y-0.5)/(x-(-1.5)) = (1/sqrt(2))
EDIT:That was really cool\pi is named pi not pie