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

If this part uses an earlier result, revisit (ii) Vertical tangents or open all working together.

(iii)

Normals

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

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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

Open interactive exploration

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.