Parametrics problem (1 Viewer)

swizel03 · Dec 13, 2016

Hi guys!

I have been stuck on this question for ages and would really appreciate it if you help me out.

Normals to the parabola x = 2at, y = at^2 from points P(2ap, ap^2) and Q(2aq, aq^2) intersect at N. Find the equation of the locus of N if PQ passes through the point (0, 3a).

Thanks guys!

jathu123 · Dec 13, 2016

swizel03 · Dec 13, 2016

jathu123 said:
$$\noindent$ $ Normal at P: $x+py=ap^3+2ap \\ $ Normal at Q: $ x+qy=aq^3+2aq. \\$ To find the intersection of these normals, solve them simultaneously. You can subtract them to eliminate $x. \\ \therefore py-qy=ap^3-aq^3+2ap-2aq. \\y(p-q)=a(p^3-q^3)+2a(p-q)$ factorise the difference of cubes aswell $\\ y(p-q)=a(p-q)(p^2+pq+q^2)+2a(p-q)\\$ Now factorise the whole $a(p-q)$ term in the RHS$ \\ y(p-q)=a(p-q)(p^2+pq+q^2+2)\\$ Now cancel out the $(p-q)$ to find y$ \\ y=a(p^2+pq+q^2+2)\\ $ You can sub this into either of our first two equations to find $x\\$ After simplifying, the coordinate of the intersection of the normals will be $ \\N\left ( -apq(p+q),a(p^2+pq+q^2+2) \right )\\ \\$ Now we need to find a relationship between $p$ and $q$ to find the locus of N. We can use the fact that $PQ$ passes through $(0,3a)$ to find this relationship. \\$

$$equation of PQ: $ y=\frac{1}{2}x(p+q)-apq.$ The point $(0,3a)$ must satisfy this equation since the chord passes through this. So sub this in $\\ 3a=\frac{1}{2}\cdot 0\cdot (p+q)-apq \\ 3a+apq=0 \\ a(3+pq)=0 \\ a\neq 0,\Rightarrow pq=-3\\ $ We can use this fact, that $pq=-3$ to find the locus of N$\\ x=-apq(p+q) \\ = 3a(p+q)\\\therefore (p+q)=\frac{x}{3a}\Rightarrow (p+q)^2=\frac{x^2}{9a^2}$ ------ $ (1) \\ \\ y=a(p^2+pq+q^2+2) \\ a((p+q)^2-pq+2)$ Now sub in $pq=-3$ and equation (1) $\\ y=a\left ( \frac{x^2}{9a^2} +5\right )\\ \\ \therefore x^2=9a(y-5a)$

Thank You!

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Parametrics problem (1 Viewer)

swizel03

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jathu123

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swizel03

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