An ellipse becomes a circle · source reference

Printed question

The ellipse $C_1$ with the equation $\frac{x^2}{6^2}+\frac{y^2}{3^2}=1$ 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 $C_2$.

(i) Find the equation of $C_2$ in terms of $k,m,n$. [3]

(ii) Given that $C_2$ is a circle centred at $(4,-7)$, find the values of $k,m,n$. [2]

[2015/SAJC/Prelim/I/5]

Printed working

(i) Transformation 1: Stretch with scale factor $k$ parallel to $x$-axis.

Transformation 2: Translate $m$ units in positive $x$-direction.

Transformation 3: Translate $n$ units in negative $y$-direction.

$C_1:\frac{x^2}{6^2}+\frac{y^2}{3^2}=1\xrightarrow{\text{Trans 1}}\frac{(x/k)^2}{6^2}+\frac{y^2}{3^2}=1$.

$\xrightarrow{\text{Trans 2}}\frac{(x-m)^2}{(6k)^2}+\frac{y^2}{3^2}=1$.

$\xrightarrow{\text{Trans 3}}\frac{(x-m)^2}{(6k)^2}+\frac{(y+n)^2}{3^2}=1$.

$C_2:\frac{(x-m)^2}{(6k)^2}+\frac{(y+n)^2}{3^2}=1$.

(ii) If $C_2$ is a circle, then $6k=3\Rightarrow k=\frac12$.

$C_2$ is a circle centred at $(4,-7)\Rightarrow m=4,n=7$.

Teacher notes

Transformation 1: Replace $x$ with $\frac xk$.

Transformation 2: Replace $x$ with $x-m$.

Transformation 3: Replace $y$ with $y+n$.

For circle, coefficients of $x^2,y^2$ are the same.

Printed answer key

(i) $C_2:\frac{(x-m)^2}{(6k)^2}+\frac{(y+n)^2}{3^2}=1$. (ii) $k=\frac12,m=4,n=7$.

Source provenance

2025 TJC H2 Math (9758) JC2 MSM: GRAPHS AND CURVE SKETCHING. Question/answer booklet printed pages 1–14; solutions printed pages 1–28. Running metadata and repeated headings are transcribed in the topic metadata article.