1. ## elliptical question

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 - but I thought the parametric form was simply an alternate formulation of the ellipse and therefore there should be a way of solving the question without using the parametric form. Is this right?

So comparing $\displaystyle x^2 + 4y^2 = 1$ with $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ we have a = 1 and b = $\displaystyle \frac{1}{2}$

$\displaystyle P(x = a.cos \theta, y= a.sin \theta)$ and N(2,y) so the mid point is $\displaystyle M(\frac{a.cos \theta + 2)}{2}, a.sin \theta)$

therefore $\displaystyle 2x = a cos \theta + 2$
$\displaystyle 4x^2 = (a.cos \theta + 2)^2$

substitute:
$\displaystyle y = a sin \theta$
$\displaystyle 2x - 2 = a. cos \theta$

I get: $\displaystyle 4x^2 + y^2 - 8x + 3 = 0$
My maths book gives: $\displaystyle 4x^2 + 4y^2 - 8x + 3 = 0$

regards

2. 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: $\displaystyle (x = a\cdot cos \theta, y= \bold{\color{red}b} \cdot sin \theta)$.
If you use this equation and the values for a and b you already know you'll get the equation which is given by your textbook.

The equation describes a semi-circle around (1, 0) with radius r = 0.5.

3. Yes, silly mistake. Thanks for finding it. But why do you say the locus is a semi circle? Isn't it a circle?

4. Originally Posted by s_ingram
Yes, silly mistake. Thanks for finding it. But why do you say the locus is a semi circle? Isn't it a circle?
Correct, you are right.
I split the point P into 2 points A and B to construct the locus: The point D (in red) is the midpoint of AN and the point C (in blue) is the midpoint of BN.

Then you'll get two semi-circles which form a complete circle.