Vertical tangents on a tilted ellipse

The equation of a closed curve is given by (nx+y)2+4(nx−y)2=20(nx+y)^2+4(nx-y)^2=20, where nn is a constant. Find

(i) the gradient of the curve at the point (x,y)(x,y), marks

(ii) the coordinates of the points on the curve at which the tangent is parallel to the yy-axis, marks

(iii) the equations of the normal at the points where the curve cuts the xx-axis. marks

[2009/IJC/Prelim/I/11]

Total marks: 10

Solution

(i)

Implicit differentiation

The closed curve (nx+y)2+4(nx−y)2=20(nx+y)^2+4(nx-y)^2=20, where nn is a constant. Find the gradient of the curve at the point (x,y)(x,y).

Hint

  1. Differentiate both sides with respect to xx using the chain rule.

  2. Collect dydx\frac{dy}{dx} and solve.

Answer · (i)

dy/dx=n(3y−5nx)/(5y−3nx)dy/dx=n(3y-5nx)/(5y-3nx).

Closed-curve assumption requires n≠0.

(ii)

Vertical tangents

Find the coordinates of the points on the curve at which the tangent is parallel to the yy-axis.

Explore this idea

Hint

  1. Vertical tangent: dydx\frac{dy}{dx} undefined, so the denominator is 0.

  2. Combine this with the equation of the curve.

Answer · (ii)

(5/(2n),3/2)(5/(2n),3/2) and (−5/(2n),−3/2)(-5/(2n),-3/2).

Explore the idea

Where does the tilted ellipse have a vertical tangent?

Shown for n = 1. Jump to the two special points.

Position along the curve(0, 2)
Try an example
Tangent gradient
0.6
Normal gradient
-1.66667

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.

  • Ellipse
  • Tangent
  • Normal
  • A right angle between the tangent and normal when the guide is enabled
Key idea

The ellipse is symmetric about the origin, so vertical tangents come in opposite pairs.

Ideas to try
  1. Where does the tangent become horizontal? (Hint: the numerator.)
  2. Why are the two vertical-tangent points opposite each other?
(iii)

Normals

Find the equations of the normals at the points where the curve cuts the xx-axis.

Explore this idea

Hint

  1. Put y=0y=0 in the curve to find the points.

  2. Find the gradient there, then the normal gradient −1m-\frac1m.

Answer · (iii)

y=−3x/(5n)±6/(5n2)y=-3x/(5n)\pm6/(5n^2).

Both x-axis intersections require a normal; n≠0.

Explore the idea

Parallel normals at the x-intercepts

Jump to each x-intercept. The tangent gradient is 5/3 at both, so the normals are parallel.

Position along the curve(2, 0)
Try an example
Tangent gradient
1.66667
Normal gradient
-0.6

At (2n,0)\left(\frac2n,0\right): tangent gradient 5n3\frac{5n}3, normal gradient −35n-\frac3{5n}.

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

  • Ellipse
  • Tangent
  • Normal
  • A right angle between the tangent and normal when the guide is enabled
Key idea

Points related by the symmetry (x,y)→(−x,−y)(x,y)\to(-x,-y) have equal tangent gradients.

Ideas to try
Compare the gradient readouts at the two intercepts.