A normal intersection and a midpoint locus

A curve is defined parametrically by x=att+1x=\frac{at}{t+1}, y=at2t+1y=\frac{at^2}{t+1}, where aa is a non-zero constant and t∈Rt\in\mathbb R, t≠−1t\ne-1.

(i) Show that the equation of the normal to the curve at the point TT with parameter tt is given by t(t+1)(t+2)y+(t+1)x=at(t3+2t2+1)t(t+1)(t+2)y+(t+1)x=at(t^3+2t^2+1). marks

(ii) The normal to the curve at the point P(−a,12a)P(-a,\frac12a) meets the curve again at point QQ. Find the exact coordinates of QQ in terms of aa. marks

(iii) Let MM be the mid-point of PQPQ. Find a cartesian equation of the curve traced by MM if aa varies. marks

[2022/NYJC/Prelim/I/7]

Total marks: 10

Solution

(i)

Normal to a parametric curve

x=att+1x=\dfrac{at}{t+1}, y=at2t+1y=\dfrac{at^2}{t+1}. Show that the normal at the point TT with parameter tt is t(t+1)(t+2)y+(t+1)x=at(t3+2t2+1)t(t+1)(t+2)y+(t+1)x=at(t^3+2t^2+1).

Explore this idea

Hint

  1. Differentiate xx and yy with respect to tt (quotient rule).

  2. Find dydx\frac{dy}{dx}, then the normal gradient.

  3. Write the normal and clear fractions.

Answer · (i)

t(t+1)(t+2)y+(t+1)x=at(t3+2t2+1)t(t+1)(t+2)y+(t+1)x=at(t^3+2t^2+1).

a≠0,t≠−1a\ne0,t\ne-1; formula also covers horizontal-tangent parameters 0 and −2.

Explore the idea

A tangent and a normal at every parameter (a = 1)

Slide the point and watch the normal turn. The final formula covers the special parameters t = 0 and t = −2 too.

Position along the curve(0.5, 0.5)t = 1
Try an example
Tangent gradient
3
Normal gradient
-0.33333

The tangent follows the curve at the selected point. Its normal is perpendicular to it.

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • Curve (t > −1 branch)
  • Tangent
  • Normal
  • A right angle between the tangent and normal when the guide is enabled
Key idea

The same formula for the normal works at every parameter, even where the tangent is horizontal.

Ideas to try
Where is the tangent horizontal? What happens to the normal there?
(ii)

Intersecting a normal with the curve

The normal at P(−a,12a)P\left(-a,\tfrac12a\right) meets the curve again at QQ. Find the exact coordinates of QQ in terms of aa.

Explore this idea

Hint

  1. Find the parameter at PP: att+1=−a\frac{at}{t+1}=-a.

  2. Put it into the normal formula from (i).

  3. Substitute the parametric equations into that line; one root is PP.

Answer · (ii)

Q=(11a/14,121a/42)Q=(11a/14,121a/42).

Explore the idea

The normal at P meets the curve again at Q

Shown for a = 1. Jump to P, then to Q.

Position along the curve(-1, 0.5)t = -0.5
Try an example
Tangent gradient
-0.75

PP: parameter t=−12t=-\frac12, tangent gradient −34-\frac34. The purple line is its normal y=43x+116y=\frac43x+\frac{11}6.

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • Curve for a = 1
  • Tangent at the moving point
Key idea

Substituting a curve into a line gives a polynomial; the point you started from is always a root.

Ideas to try
Is QQ where you would expect the purple line to cross the curve again?
(iii)

Locus by eliminating a parameter

Let MM be the mid-point of PQPQ. Find a cartesian equation of the curve traced by MM as aa varies.

Explore this idea

Hint

  1. P=(−a,a2)P=\left(-a,\frac a2\right) and Q=(11a14,121a42)Q=\left(\frac{11a}{14},\frac{121a}{42}\right), so both coordinates of MM are multiples of aa.

  2. Eliminate aa.

Answer · (iii)

y=−142x/9y=-142x/9, excluding the origin.

a≠0.

Explore the idea

The midpoint moves along a line

Switch on the midpoints for a = 1, 2 and −1: all lie on the same line.

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

Ideas to try
What happens to MM as a→0a\to0?