1. ## Gradient of the tangent to the ellipse

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?

2. Originally Posted by Arron
I am stuck on this question.
x = root 3 cos t, y = sin t (t\in [0,2pie])
point P is where t = 5pie/6, ie (-1.5,0.5)
Can you find each of these
The evaluate at the given point.
How do you use those to find the gradient?

BTW: $\pi$ is named pi not pie.

3. 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:
\pi is named pi not pie
That was really cool

4. Thank you so much for your help. Finally could you please let me know how to find the point where the tangent meets the x-axis.

5. Originally Posted by Arron
Thank you so much for your help. Finally could you please let me know how to find the point where the tangent meets the x-axis.
You have the equation of the line, when the line reaches the x axis the value of y is zero. Solve for x.

-Dan

6. Originally Posted by Arron
Thank you so much for your help. Finally could you please let me know how to find the point where the tangent meets the x-axis.
BUT be careful.

7. Originally Posted by Plato
BUT be careful.
Why?

8. Originally Posted by Sambit
Why?
The slope is not correct.
What is $\dfrac{dy}{dx}\left(\frac{5\pi}{6}\right)~?$

9. The denominator should be -root(3) sin(5pi/6), not root (3 sin (5pi/6))
And, of course, the derivative of sin(t) is cos(t) not -cos(t) but sambit may have just moved the - from the denominator to the numerator,

10. Ok I got it. Sorry for that.

11. Guys are you saying that the correct answer is cos(5pi/6)/-Root 3 sin (5pi/6) = 1/Root 2.

12. I am still not sure about the gradient, am I correct in the last post?