All ellipse normals through a fixed point

MI Prelim 9758/2023/02/Q4. The curve CC has equation 2x2+y2=202x^2+y^2=20.

(i) Sketch curve CC, giving the exact coordinates of any points where the curve meets the axes. marks

(ii) Show that dydx=−2xy\frac{dy}{dx}=-\frac{2x}{y}. mark

(iii) It is given that point P(a,b)P(a,b) is on CC. The normal to CC at PP passes through the point (1,0)(1,0). Find the possible coordinates of PP. marks

[2023/MI/Prelim/II/4]

Total marks: 8

Solution

(i)

Sketching an ellipse

Sketch the curve CC: 2x2+y2=202x^2+y^2=20, giving the exact coordinates of any points where it meets the axes.

Explore this idea

Hint

  1. Divide by 20 to get the standard form.

  2. Read off the intercepts.

Ellipse 2x²+y²=20
Ellipse 2x²+y²=20Scroll across to see the full diagram.

Answer · (i)

Ellipse: intercepts (±10,0),(0,±25)(\pm\sqrt{10},0),(0,\pm2\sqrt5).

Explore the idea

A tall ellipse

Switch on the intercepts.

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
Which intercepts are further from the origin, the xx or the yy ones?
(ii)

Implicit differentiation

Show that dydx=−2xy\dfrac{dy}{dx}=-\dfrac{2x}y.

Hint

  1. Differentiate each term with respect to xx.

Answer · (ii)

dy/dx=−2x/ydy/dx=-2x/y.

y≠0y\ne0.

(iii)

Normals through a point

P(a,b)P(a,b) is on CC and the normal at PP passes through (1,0)(1,0). Find the possible coordinates of PP.

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Hint

  1. Normal gradient at PP is b2a\frac b{2a}.

  2. The normal passes through (1,0)(1,0), so its gradient is also b−0a−1\frac{b-0}{a-1}.

  3. Solve the condition, **without dividing by bb**, and use the curve.

Answer · (iii)

P=(±10,0)P=(\pm\sqrt{10},0) or (−1,±32)(-1,\pm3\sqrt2).

Retain both axis-point normals; dividing by b would lose them.

Explore the idea

Which normals pass through (1, 0)?

Jump to each candidate. The purple normal runs through the marked point (1, 0) exactly at these four points.

Position along the curve(-1, 4.24264)
Try an example
Tangent gradient
0.4714
Normal gradient
-2.12132

At (−1,32)(-1,3\sqrt2) the normal has gradient b2a=−322\frac b{2a}=-\frac{3\sqrt2}2 and passes through (1,0)(1,0).

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

  • 2x² + y² = 20
  • Tangent
  • Normal
  • A right angle between the tangent and normal when the guide is enabled
Key idea

Exactly four normals pass through (1,0)(1,0): two from the xx-axis points (b=0b=0) and two from a=−1a=-1.

Ideas to try
  1. Slide to a point on the top of the ellipse: does the normal go through (1,0)(1,0)?
  2. Which preset points lie on the xx-axis?