An ellipse becomes a circle

C_1

Total marks: 5

Solution

(i)

Transforming an equation

The ellipse $C_1$ with equation $\dfrac{x^2}{6^2}+\dfrac{y^2}{3^2}=1$ is transformed by a stretch with scale factor $k$ parallel to the $x$ -axis, followed by a translation of $m$ units in the positive $x$ -direction, followed by a translation of $n$ units in the negative $y$ -direction, to form $C_2$ . Find the equation of $C_2$ in terms of $k$ , $m$ and $n$ .

Hint

Express each transformation as a replacement: stretch parallel to the $x$ -axis replaces $x$ by $\frac xk$ ; a translation right by $m$ replaces $x$ by $x-m$ ; down by $n$ replaces $y$ by $y+n$ .
Apply them in order, simplifying after each.

Answer · (i)

\frac{(x-m)^2}{(6k)^2}+\frac{(y+n)^2}{9}=1

(ii)

Circles from ellipses

Given that $C_2$ is a circle with centre $(4,-7)$ , find the values of $k$ , $m$ and $n$ .

Explore this idea

Hint

A circle needs equal semi-axes. Compare $6k$ with $3$ .
The centre of $C_2$ is $(m,-n)$ .

Answer · (ii)

k=1/2,m=4,n=7

Explore the idea

Squash, slide right, slide down

Three steps turn the wide ellipse into a circle of radius 3 centred at (4, −7).

Start: C₁: x²/36 + y²/9 = 1

The original ellipse: semi-axes 6 and 3, centred at the origin.

Marked points at this step, rounded to 6 significant figures.
Point	x	y
(0,0)	0	0

Shape at the selected step
Centre of the current shape

Ideas to try

Which step makes the ellipse round? Why does only $k=\frac12$ work?
Check the equation of the final circle against the centre $(4,-7)$ .

An ellipse becomes a circle

Solution

Transforming an equation

Stretch parallel to the $x$ -axis

Translate $m$ units right

Translate $n$ units down

Circles from ellipses

Equal semi-axes

Read the centre