Originally Posted by

**s_ingram** hi guys,

I have the following problem:

The point P lies on the ellipse $\displaystyle x^2 + 4y^2 = 1$ and N is the foot of the perpendicular from P to the line x = 2 (see attachment). Find the equation of the mid point of PN as P moves on the ellipse.

My answer does not agree with the one in my maths book, so I would just like to confirm whether the book is correct or not. Secondly, I have solved this problem using the parametric form of the point $\displaystyle (x = a.cos \theta, y= a.sin \theta)$ and as far as I can see, it is the only way to solve this problem - **<<<<<< that is correct but the parametric equation of an ellipse is: $\displaystyle (x = a\cdot cos \theta, y= \bold{\color{red}b} \cdot sin \theta)$.**