A family of lines can miss a rational curve · source reference

Printed question

The curve $C$ has equation $y=\frac{x^2+b}{x+a}$, where $a>0,b>0$.

(i) State the coordinates of the intersection(s) of the curve $C$ with the axes in terms of $a,b$. [1]

(ii) Find the equation(s) of the asymptote(s). [2]

(iii) Draw a sketch of the curve $C$, labelling the equation(s) of its asymptote(s) and coordinates of any intersection with the axes. [2]

(iv) Hence find the range of values of $k$, where $k$ is a positive constant, for which the equation $x^2+b=(x+a)(kx-a)$ has no real root. [2]

[2009/SRJC/Prelim/I/7]

Printed working

(i) $(0,\frac ba)$.

(ii) $y=\frac{x^2+b}{x+a}=(x-a)+\frac{b+a^2}{x+a}$.

$x=-a$; $y=x-a$ are equations of the asymptotes.

(iii) Printed graph below.

(iv) $x^2+b=(x+a)(kx-a)\Rightarrow\frac{x^2+b}{x+a}=kx-a$.

From graph, $0<k\le1$. (Note: $k>0$).

Teacher notes

For eqn to have no real roots, the line $l:y=kx-a$, where $k$ is the gradient and which passes through pt $(0,-a)$, does not intersect the curve $y=\frac{x^2+b}{x+a}$.

Note gradient of oblique asymptote of curve is 1 while gradient of horizontal line through pt $(0,-a)$ is zero. Any line $l$ between the above 2 lines does not intersect the curve.

Printed answer key

(i) $(0,\frac ba)$. (ii) Asymptotes $x=-a,y=x-a$. (iii) Native graph above. (iv) $0<k\le1$.

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.