A normal intersection and a midpoint locus · source reference

Printed question

A curve is defined parametrically by $x=\frac{at}{t+1}$, $y=\frac{at^2}{t+1}$, where $a$ is a non-zero constant and $t\in\mathbb R$, $t\ne-1$.

(i) Show that the equation of the normal to the curve at the point $T$ with parameter $t$ is given by $t(t+1)(t+2)y+(t+1)x=at(t^3+2t^2+1)$. [4]

(ii) The normal to the curve at the point $P(-a,\frac12a)$ meets the curve again at point $Q$. Find the exact coordinates of $Q$ in terms of $a$. [4]

(iii) Let $M$ be the mid-point of $PQ$. Find a cartesian equation of the curve traced by $M$ if $a$ varies. [2]

[2022/NYJC/Prelim/I/7]

Printed working

(i) Curve equation (1): $x=at/(t+1)$, $y=at^2/(t+1)$.

$$\frac{dx}{dt}=\frac{a(t+1)-at}{(t+1)^2}=\frac a{(t+1)^2},\quad\frac{dy}{dt}=\frac{2at(t+1)-at^2}{(t+1)^2}=\frac{at^2+2at}{(t+1)^2}.$$

Thus $dy/dx=t^2+2t$.

Normal at $T$: $y-\frac{at^2}{t+1}=-\frac1{t(t+2)}[x-\frac{at}{t+1}]$.

$$t(t+2)y-\frac{at^2}{t+1}t(t+2)=-x+\frac{at}{t+1},$$ $$t(t+1)(t+2)y-at^3(t+2)=-(t+1)x+at,$$ $$t(t+1)(t+2)y+(t+1)x=at(t^3+2t^2+1)$$

(shown). At $t=0,-2$ the first slope form is undefined; the final equation correctly gives the vertical normal.

(ii) At $P$, $at/(t+1)=-a\Rightarrow t=-1/2$. Substitute into (i): $-\frac38y+\frac12x=-\frac{11}{16}a\Rightarrow-6y+8x=-11a$ (2).

Alternatively at $t=-1/2$, $x=-a$, $y=a/2$, $dy/dx=-3/4$, so $y-a/2=\frac43(x+a)\Rightarrow y=\frac43x+\frac{11}6a$.

Substitute (1) into (2):

$$-6\frac{at^2}{t+1}+8\frac{at}{t+1}=-11a\Rightarrow6t^2-8t=11(t+1)\Rightarrow6t^2-19t-11=0.$$

$(2t+1)(3t-11)=0\Rightarrow t=-1/2$ or $11/3$. For $Q$, $t=11/3$,

$$x=\frac{a(11/3)}{11/3+1}=\frac{11}{14}a,\qquad y=\frac{a(11/3)^2}{11/3+1}=\frac{121}{42}a.$$

(iii) Midpoint $M=(-\frac3{28}a,\frac{71}{42}a)$. Thus $x=-3a/28$, $y=71a/42$ ⇒ $y=-\frac{142}9x$ (Cartesian equation). Since $a\ne0$, the origin is excluded from the locus.

Teacher notes

Use $dy/dx=(dy/dt)/(dx/dt)$.

Gradient of normal $=-1/\text{gradient of tangent}=-1/(dy/dx)$ where finite.

Substitute $t=-1/2$ into the normal equation from (i). Substitute equation (1) into equation (2) to find its intersections with the curve.

To obtain a Cartesian equation, an equation involving $x$ and $y$ only, eliminate variable $a$ from the midpoint equations.

Answer key

(ii) $(11a/14,121a/42)$. (iii) $y=-142x/9$.