Hi there MHF,

After an hour of intensive algebra, I am still unable to solve this question.

Could someone please help me with Part A). And also, if they are kind enough, Part B).

Thank you.

http://img545.imageshack.us/img545/858/locus.png

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- Nov 29th 2011, 08:39 PMeskimogeniusLocus and the Parabola Problem
Hi there MHF,

After an hour of intensive algebra, I am still unable to solve this question.

Could someone please help me with Part A). And also, if they are kind enough, Part B).

Thank you.

http://img545.imageshack.us/img545/858/locus.png - Nov 29th 2011, 11:25 PMearbothRe: Locus and the Parabola Problem
1. You probably have determined the coordinates of the focus as S(0, 1).

2. The equation of the parabola is $\displaystyle y=\frac14 \cdot x^2$.

That means the slope of the tangent to the parabola is calculated by $\displaystyle y'=\frac12 \cdot x$

3. Let $\displaystyle P\left(p,\frac14p^2 \right)$ be a point on the parabola. The tangent $\displaystyle t_P$ at P to the parabola has the equation:

$\displaystyle y-\frac14p^2=\frac12p \cdot (x-p)$

$\displaystyle t_P: y=\frac12p \cdot x - \frac14p^2$

4. The normal $\displaystyle n_P$ in P has the slope $\displaystyle m = -\frac2p$. Since the normal passes through P too it has the equation:

$\displaystyle y-\frac14p^2=-\frac2p \cdot (x-p)$

$\displaystyle y = -\frac2p \cdot x +2+\frac14p^2$

5. Now determine the point of intersection of the parabola and the normal.

6. Show that $\displaystyle \overrightarrow{SP} \cdot \overrightarrow{SQ} = 0$ that means the angle $\displaystyle \angle(PSQ) = 90^\circ$. - Nov 29th 2011, 11:49 PMeskimogeniusRe: Locus and the Parabola Problem
Having immense troubles here trying to determine the point of intersection of the parabola and the normal.

When you substitute it in, you have on one side, x^2, whilst on the other side, x. It is impossible to make x the subject of the equation.

How do I complete your step 5?

In other attempts, I have tried to use a separate variable for Q, with P being (2ap, ap^2) and Q being (2aq, aq^2). And then using Pythagoras' Theorem to prove true for the right angle. However, the algebra becomes intense and you are unable to eliminate and cancel down to the RHS. Also, I have attempted to use the gradients, then attempting to substitute the given condition of QS = 2PS, however, that fails also with the algebra coming up to powers of 4. - Nov 30th 2011, 01:09 AMearbothRe: Locus and the Parabola Problem
The intersection of the parabola and the normal:

equation of parabola: $\displaystyle y = \frac14 x^2$

equation of normal: $\displaystyle y = -\frac2p x+2+\frac14p^2$

Solve for x:

$\displaystyle \frac14 x^2 = -\frac2p x + 2 + \frac14p^2$

$\displaystyle x^2+\frac8p x - 8 - p^2=0$

Apply the quadratic formula:

$\displaystyle x = \frac{-\frac8p \pm \sqrt{\frac{64}{p^2} - 4(-8-p^2)}}2$

$\displaystyle x = \frac{-\frac8p \pm \frac2p \cdot \sqrt{p^4+8p^2+16 }}2$

The radicand is a complete square!

Therefore you'll get:

$\displaystyle \underbrace{x = p}_{P} ~\lor ~ \underbrace{x = -p-\frac8p}_{Q}$

So Q has the coordinates $\displaystyle Q\left(-p-\frac8p, \frac14\left(p+\frac8p \right)^2\right)$

To prove the orthogonality of $\displaystyle \overline{SP}$ and $\displaystyle \overline{SQ}$ respectively I would use vectors as I've described in my previous post.