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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: 24221 on August 06, 2018, 11:25:48 am

Title: Re: Parametric, Cambridge Ext 1, Exercise 9E Q12
Post by: 24221 on August 06, 2018, 11:25:48 am: A focal chord AB of a parabola meets the directrix at D. Prove the focus S divides AB internally in the ratio k:1. Then D divides AB externally in the ratio k:1

Thank you, :D
Title: Re: Re: Parametric, Cambridge Ext 1, Exercise 9E Q12
Post by: RuiAce on August 07, 2018, 09:29:57 am: Quote from: 24221 on August 06, 2018, 11:25:48 am
A focal chord AB of a parabola meets the directrix at D. Prove that if focus S divides AB internally in the ratio k:1. Then D divides AB externally in the ratio k:1

Thank you, :D
I was able to retrace the question with thanks to the textbook reference. In the future please be cautious of typos because not everyone will have the textbook and thus the question won't make any sense to them.
$\text{One may take }A\text{ to be }(2ap,ap^2)\text{ and }B\text{ to be }(2aq,aq^2)\\ \text{however we're told that }AB\text{ is a focal chord, so }pq = -1.\\ \text{Therefore, consider the points }\boxed{A\left( 2ap, ap^2\right)}\text{ and }\boxed{B\left( -\frac{2a}{p}, \frac{a}{p^2} \right)}$
Remark: One may show that $D$ is the point $ \left( \frac{4a}{p^{-1} - p}, -a \right) $, but we only really care about the $y$-coordinate here.
$\text{Suppose that }S(0,a)\text{ divides }AB\text{ internally in the ratio }k:1.\\ \text{Applying the ratio division formula to the }x\text{-coordinates,}\\ \begin{align*} 0 &= \frac{-\frac{2ak}{p} + 2ap}{k+1}\\ \therefore 0 &= -\frac{k}{p} + p \end{align*}\\ \text{so we see that in fact, }\boxed{k= p^2}.$
One may check that $k=p^2$ actually works by testing it out with the corresponding $y$-coordinates as well.
$\text{We want to ensure that }D\text{ divides }AB\text{ in the ratio }-k:1.\\ \text{ Using the formula on the }y\text{-coordinates and the fact that }k=p^2,\\ \begin{align*}y_D &= \frac{-\frac{ak}{p^2} + ap^2}{-k+1}\\ &= \frac{a-ap^2}{-p^2+1} \\ &= -a\end{align*}\\ \text{so }D\text{ must lie on the directrix }y=-a\text{ as required.}$
Title: Re: Parametric, Cambridge Ext 1, Exercise 9E Q12
Post by: 24221 on August 07, 2018, 07:35:12 pm: Thank you very much, this helped out a lot!