Respect both square-root domains · source reference

Printed question

The hyperbola $C$ passes through the point $(2,0)$ and has oblique asymptotes $y=-2x$ and $y=2x$.

(i) Sketch $C$, showing the relevant features of the curve. [2]

(ii) Write down the equation of $C$. [1]

(iii) By adding a suitable curve to your sketch in part (i), solve the inequality $\sqrt{\frac{x^2}{4}-1}<\sqrt{x-1}$. [3]

[2019/TJC/Prelim/I/3]

Printed working

(i) Printed sketch below: vertices $(-2,0),(2,0)$, asymptotes $y=-2x,y=2x$.

(ii) The hyperbola is of the form $\frac{x^2}{2^2}-\frac{y^2}{b^2}=1$.

Oblique asymptotes is $y=\pm\frac b2x=\pm2x$.

$\therefore\frac b2=2\Rightarrow b=4$.

Hence equation of hyperbola is $\frac{x^2}{4}-\frac{y^2}{16}=1$.

(iii) $\frac{y^2}{16}=\frac{x^2}{4}-1\Rightarrow y^2=16(\frac{x^2}{4}-1)\Rightarrow y=\pm4\sqrt{\frac{x^2}{4}-1}$.

$4\sqrt{\frac{x^2}{4}-1}<4\sqrt{x-1}$.

Add sketch of curve $y=4\sqrt{x-1}$ to diagram in (i).

The 2 curves intersect at $x=4$.

Hence for $\sqrt{\frac{x^2}{4}-1}<\sqrt{x-1}$, $2\le x<4$.

Teacher notes

The centre of hyperbola $C$ (which is the pt of intersection of the asymptotes) is the origin. So $(2,0)$ is a vertex of $C$ as it lies on the line of symmetry $y=0$. Thus the other vertex is $(-2,0)$.

As the distance of the vertices from the centre of $C$ is 2 along the $x$-direction, hyperbola is of form $\frac{x^2}{2^2}-\frac{y^2}{b^2}=1$. Note when $y=0$, $x=\pm2$.

To obtain oblique asymptotes, let $\frac{x^2}{2^2}-\frac{y^2}{b^2}=0$, i.e. $\frac{y^2}{b^2}=\frac{x^2}{2^2}$.

Note that $4\sqrt{x-1}\ge0$ and $4\sqrt{\frac{x^2}{4}-1}\ge0$.

Printed answer key

(i) Native graph above. (ii) Equation of $C$: $\frac{x^2}{4}-\frac{y^2}{16}=1$. (iii) $2\le x<4$.

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.