A vertical tangent forms a square

The equation of a curve is given by 4x3+3x2y=y3−24x^3+3x^2y=y^3-2. Find dydx\frac{dy}{dx} in terms of xx and yy, simplifying your answer. marks

The curve meets the line y=−xy=-x at point PP. Find (i) the coordinates of PP marks and (ii) the equation of the tangent at PP. marks

The tangent to the curve at PP cuts the xx-axis at RR and the normal to the curve at PP cuts the yy-axis at QQ. If OO denotes the origin, use the results above to give a geometrical description of the quadrilateral OQPROQPR. mark

[2012/NYJC/Prelim/I/3]

Total marks: 7

Solution

Derivative

Implicit differentiation

The curve has equation 4x3+3x2y=y3−24x^3+3x^2y=y^3-2. Find dydx\dfrac{dy}{dx} in terms of xx and yy, simplifying your answer.

Hint

  1. Differentiate every term with respect to xx, using the product rule on 3x2y3x^2y and the chain rule on y3y^3.

  2. Collect the dydx\frac{dy}{dx} terms on one side and factorise.

Answer · Derivative

dy/dx=2x(2x+y)/[(y−x)(y+x)]dy/dx=2x(2x+y)/[(y-x)(y+x)].

At regular points where denominator is nonzero.

(i)

Intersecting a curve and a line

The curve meets the line y=−xy=-x at the point PP. Find the coordinates of PP.

Hint

  1. Substitute y=−xy=-x into the equation of the curve and solve for xx.

Answer · (i)

P=(−1,1)P=(-1,1).

(ii)

Tangent from implicit differentiation

Find the equation of the tangent at PP.

Explore this idea

Hint

  1. Substitute P(−1,1)P(-1,1) into dydx\frac{dy}{dx}.

  2. Look at the denominator.

Answer · (ii)

Tangent x=−1x=-1.

Explore the idea

The tangent turns vertical at P

Drag the point along the curve, or jump to P. The amber line is the tangent and the purple line is the normal.

Position along the curve(-0.70202, -0.50252)
Try an example
Tangent gradient
-11.13967
Normal gradient
0.08977

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.

  • 4x³ + 3x²y = y³ − 2
  • Tangent
  • Normal
  • A right angle between the tangent and normal when the guide is enabled
Key idea

A zero denominator with a non-zero numerator means a vertical tangent: the equation is x=x= constant.

Ideas to try
  1. Slide towards PP from either side and watch the gradient grow.
  2. At PP, what are the directions of the tangent and the normal?
Quadrilateral

Geometric interpretation

The tangent at PP cuts the xx-axis at RR and the normal at PP cuts the yy-axis at QQ. Describe the quadrilateral OQPROQPR.

Explore this idea

Hint

  1. The tangent is x=−1x=-1 and the normal is perpendicular to it, so y=1y=1.

  2. Find RR and QQ and look at the four vertices.

Answer · Quadrilateral

OQPROQPR is a unit square; Q=(0,1),R=(−1,0)Q=(0,1),R=(-1,0).

Explore the idea

The tangent and normal at P frame a unit square

The shaded corner points are O, Q, P and R.

Position along the curve(-1, 1)
Try an example
Tangent gradient
vertical
Normal gradient
0

The tangent x=−1x=-1 and the normal y=1y=1 together with the axes make the square OQPROQPR with side 1.

The tangent is vertical and the normal is horizontal. Both axes use equal scales.

  • Curve
  • Tangent x = −1 and its side RP
  • Normal y = 1 and its side QP
  • A right angle between the tangent and normal when the guide is enabled
Key idea

A vertical tangent and its horizontal normal meet the axes at right angles, making a square.

Ideas to try
Check the lengths OQOQ, QPQP, PRPR and RORO.