Pure Mathematics Practice: Conic Sections — Answers

Questions: Conic Sections Drills.

Question 1

(a)

$$(x-3{)}^2-9+(y+2{)}^2-4-12=0,$$

so

$$(x-3{)}^2+(y+2{)}^2=25.$$

The centre is $(3,-2)$ and the radius is $5$. [3]

(b) On the $x$-axis, $y=0$:

$$(x-3{)}^2+4=25,$$

so

$$x=3\pm \sqrt{21}.$$

The points are $(3-\sqrt{21},0)$ and $(3+\sqrt{21},0)$. [2]

(c) The lowest and highest points are $5$ units below and above the centre, so

$$-7\le y\le 3.$$

[1]

Question 2

(a) Centre $(-1,2)$; horizontal semi-axis $4$ and vertical semi-axis $3$. [2]

(b)

$$-5\le x\le 3,-1\le y\le 5.$$

[2]

(c) The sketch is centred at $(-1,2)$ and passes through

$$(-5,2),(3,2),(-1,-1),(-1,5).$$

The horizontal semi-axis is the longer one. [2]

Question 3

(a) The centre is $(2,-1)$. Since the positive squared term is the $y$-term, the branches open upwards and downwards. The vertices are

$$(2,-1\pm 2),$$

that is, $(2,1)$ and $(2,-3)$. [3]

(b) For

$$\frac{(y-k{)}^2}{a^2}-\frac{(x-h{)}^2}{b^2}=1,$$

the asymptotes are $y-k=\pm \frac{a}{b}(x-h)$. Hence

$$y+1=\pm \frac{2}{3}(x-2).$$

[2]

(c) Put $x=0$:

$$\frac{(y+1{)}^2}{4}-\frac{4}{9}=1.$$

Thus

$$(y+1{)}^2=\frac{52}{9},y=-1\pm \frac{2\sqrt{13}}{3}.$$

The points are

$$\left(0,-1+\frac{2\sqrt{13}}{3}\right)\text{and}\left(0,-1-\frac{2\sqrt{13}}{3}\right).$$

[3]

(d) The sketch has two vertical branches approaching, but never meeting, the two asymptotes. [2]

Question 4

(a)

$$(y-2{)}^2-4-8x+12=0$$

gives

$$(y-2{)}^2=8(x-1).$$

[1]

(b) The vertex is $(1,2)$, the axis is $y=2$, and $x\ge 1$. [2]

(c) The parabola opens to the right. Its only $x$-axis intercept is found by putting $y=0$:

$$4=8(x-1),$$

so it is $\left(\frac{3}{2},0\right)$. [1]